Because Numbers Matters
Numbers? Do they really matter in Chemistry? Well, probably you may think not. But after reading this article you will be clued-up on this very interesting topic regarding numbers in Chemistry.
In Chemistry, we do not literally define numbers. It simply means quantity. And quantity is an integral part for our topic today. We are going to tackle different ways of quantitative analysis using our past knowledge in Molarity, Percentage Concentration, Redox Reactions, and many more. So let us have a brief recap.
What is Molarity?. Molarity is the number of moles of solute dissolved in one liter of solution. The units, therefore are moles per liter, specifically it's moles of solute per liter of solution. You must be very careful to distinguish between moles and molarity. "Moles" measures the amount or quantity of material you have while "molarity" measures the concentration of that material. So when you're given a problem or some information that says the concentration of the solution is 0.1 M that means that it has 0.1 mole for every liter of solution. It does not mean that it is 0.1 moles. Please be sure to make that distinction.
What is Percentage composition? Percentage composition is a common way of expressing the concentration of a solution. It is a straightforward approach that you use when dealing with the composition of compounds. There are, however, some differences. You must remember that the concentrations of solution are varying while the composition of compounds is constant. Another is that the percentages can be calculated using volumes as well as weights, or even both together. One way of expressing concentrations is by volume percent. Another is by weight percent and weight/volume percent.
What is Redox reaction? Well, any chemical reaction in which the oxidation numbers of the atoms are changed is an oxidation-reduction reaction. Such reactions are also known as redox reactions, which is shorthand for reduction-oxidation reactions. Oxidation involves an increase in oxidation number, while reduction involves a decrease in oxidation number. Usually the change in oxidation number is associated with a gain or loss of electrons.
Now we have backgrounds of the different concepts involved for this article’s topic. I know you’re a bit puzzled of what is our main topic. So, for this article we will degrade informations about TITRATION.
What is Titration? Titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator that will change its color or pH Meter is used to detect the endpoint of the reaction.
To better understand this method, I will define you some words and materials to be used. First is analyte. An analyte is the substance that our interest is focused on. We aim to find its quantitative measure and we are to do that using titration. Second, we have the titrant. Titrant is the solution involved or used in a titration to determine the concentration of an unknown solution. It is usually the solution of known concentration that is delivered by a burette into a known quantity of the solution of unknown concentration. Third, we have the color indicator. We usually utilize the color indicator to indicate if our solution has reached the equivalence point by means of the color change. And last the PH meter. PH meters are used just the same way as the color indicator, except they show the numeric value of the PH levels to know if our solution has reached the equivalence point.
There are different ways for titrimetric analysis. We have the Direct titration, Indirect titration and Back titration.
Let us first start with direct titration. Direct titration is the treatment of analyte in a vessel with an appropriate titrant, the endpoint being determined instrumentally or visually with the aid of a suitable indicator. So, the titrant is contained in a burette with its accounted molarity. The endpoint is approached directly but cautiously, and finally the titrant is added dropwise in the burette in order for the final drop to not overrun the endpoint. The quantity of the substance being titrated may be calculated from the volume and the molarity of the titrant and the equivalence factor for the substance. This process may be best evaluated by the following.
aA + tT à pP
Second, we have the indirect titration. Indirect titration employs a primarily reaction wherein the analyte reacts with a reagent (added in large excess and neednot to be accurately measured); the product of this preliminary reaction is that one that reacts with the titrant. The procedure likely that of the direct titration but we must consider the preliminary reaction and the actual titration reaction. Wherein, the preliminary reaction involves the analyte and the reagent forming its product. And, the actual titration reaction makes use of the product in the preliminary reaction together with the titrant.
Preliminary reaction: aA + rR à pP
Titration reaction: pP + tT à fF
Lastly, we have the back titration. Back titration is not carried out with the solution whose concentration is required to be known or the analyte as in the case of normal or forward titration, but with the excess volume of reactant which has been left over after completing reaction with the analyte. In the process, this need the addition of a measured volume of a volumetric solution in excess amount actually needed to react with the analyte. The excess of this solution then being titrated with the second solution. The quantity of the substance being titrated may be calculated from the difference between the volume of the volumetric solution originally added, corrected by means of a back titration and the one consumed by the titrant in the back titration. It may be simplified by this process.
Addition Reaction: aA + rR à pP
Titration Reaction: rR + tT à nN
By: Ian Christopher S. Lucas
In Chemistry, we do not literally define numbers. It simply means quantity. And quantity is an integral part for our topic today. We are going to tackle different ways of quantitative analysis using our past knowledge in Molarity, Percentage Concentration, Redox Reactions, and many more. So let us have a brief recap.
What is Molarity?. Molarity is the number of moles of solute dissolved in one liter of solution. The units, therefore are moles per liter, specifically it's moles of solute per liter of solution. You must be very careful to distinguish between moles and molarity. "Moles" measures the amount or quantity of material you have while "molarity" measures the concentration of that material. So when you're given a problem or some information that says the concentration of the solution is 0.1 M that means that it has 0.1 mole for every liter of solution. It does not mean that it is 0.1 moles. Please be sure to make that distinction.
What is Percentage composition? Percentage composition is a common way of expressing the concentration of a solution. It is a straightforward approach that you use when dealing with the composition of compounds. There are, however, some differences. You must remember that the concentrations of solution are varying while the composition of compounds is constant. Another is that the percentages can be calculated using volumes as well as weights, or even both together. One way of expressing concentrations is by volume percent. Another is by weight percent and weight/volume percent.
What is Redox reaction? Well, any chemical reaction in which the oxidation numbers of the atoms are changed is an oxidation-reduction reaction. Such reactions are also known as redox reactions, which is shorthand for reduction-oxidation reactions. Oxidation involves an increase in oxidation number, while reduction involves a decrease in oxidation number. Usually the change in oxidation number is associated with a gain or loss of electrons.
Now we have backgrounds of the different concepts involved for this article’s topic. I know you’re a bit puzzled of what is our main topic. So, for this article we will degrade informations about TITRATION.
What is Titration? Titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator that will change its color or pH Meter is used to detect the endpoint of the reaction.
To better understand this method, I will define you some words and materials to be used. First is analyte. An analyte is the substance that our interest is focused on. We aim to find its quantitative measure and we are to do that using titration. Second, we have the titrant. Titrant is the solution involved or used in a titration to determine the concentration of an unknown solution. It is usually the solution of known concentration that is delivered by a burette into a known quantity of the solution of unknown concentration. Third, we have the color indicator. We usually utilize the color indicator to indicate if our solution has reached the equivalence point by means of the color change. And last the PH meter. PH meters are used just the same way as the color indicator, except they show the numeric value of the PH levels to know if our solution has reached the equivalence point.
There are different ways for titrimetric analysis. We have the Direct titration, Indirect titration and Back titration.
Let us first start with direct titration. Direct titration is the treatment of analyte in a vessel with an appropriate titrant, the endpoint being determined instrumentally or visually with the aid of a suitable indicator. So, the titrant is contained in a burette with its accounted molarity. The endpoint is approached directly but cautiously, and finally the titrant is added dropwise in the burette in order for the final drop to not overrun the endpoint. The quantity of the substance being titrated may be calculated from the volume and the molarity of the titrant and the equivalence factor for the substance. This process may be best evaluated by the following.
aA + tT à pP
Second, we have the indirect titration. Indirect titration employs a primarily reaction wherein the analyte reacts with a reagent (added in large excess and neednot to be accurately measured); the product of this preliminary reaction is that one that reacts with the titrant. The procedure likely that of the direct titration but we must consider the preliminary reaction and the actual titration reaction. Wherein, the preliminary reaction involves the analyte and the reagent forming its product. And, the actual titration reaction makes use of the product in the preliminary reaction together with the titrant.
Preliminary reaction: aA + rR à pP
Titration reaction: pP + tT à fF
Lastly, we have the back titration. Back titration is not carried out with the solution whose concentration is required to be known or the analyte as in the case of normal or forward titration, but with the excess volume of reactant which has been left over after completing reaction with the analyte. In the process, this need the addition of a measured volume of a volumetric solution in excess amount actually needed to react with the analyte. The excess of this solution then being titrated with the second solution. The quantity of the substance being titrated may be calculated from the difference between the volume of the volumetric solution originally added, corrected by means of a back titration and the one consumed by the titrant in the back titration. It may be simplified by this process.
Addition Reaction: aA + rR à pP
Titration Reaction: rR + tT à nN
By: Ian Christopher S. Lucas
Indirect Titration
Before we start discussing about Indirect Titration, let us first have a recall about Titration. Titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed.
There are three different ways in which titration is used to standardize a solution or to determine the amount of analyte present in the sample. Each method is based on the relationship between the titrant and the analyte.
* Direct Titration
* Indirect Titration or Replacement Method
* Back Titration
In this article, let’s focus only in the Indirect Titration or what others call Replacement Method. In Indirect Titration, the substance is not directly titrated but precipitated or removed from the reaction and the product formed has to be titrated. It uses a preliminary reaction wherein the analyte reacts with a reagent (added in large excess and need not be accurately measured); the product of this preliminary reaction is the one that reacts with the titrant.
Preliminary Reaction:
aA + rR pP
Titration Reaction:
pP + tT fF
Example:
A 0.2711 g sample of primary standard grade HgO was dissolved in an aqueous solution of KI. The liberated OH needed 13.22 mL of HCl to reach the end point. Calculate the molarity of HCl solution.
Preliminary Reaction: HgO + 4I- + H2O HgI4-2 + 2OH-
Titration Reaction: OH + H+ H2O
1 mol HgO 2 mol OH-
1 mol OH- 1 mol H+
To calculate the Molarity of HCl solution, first, we need to solve for the amount of mmol of it.
0.2711 g HgO x 1 mol HgO x 4 mol KI x 2 mol OH- x 1 mol HCl x 1000 mmol HCl
216.59 g HgO 1 mol HgO 4 mol KI 1 mol OH- 1 mol HCl
= 2.50 mmol HCl
And then we can now calculate the molarity of the HCl solution by dividing the mmol of HCl to the mL of HCl.
MHCl = mmol HCl
13.22 mL HCl
= 2.50 mmol HCl
13.22 ml HCl
= 0.19 M of HCl or 0.19 mmol/mL of HCl
By: Rhea Ciarina B. Ramos
There are three different ways in which titration is used to standardize a solution or to determine the amount of analyte present in the sample. Each method is based on the relationship between the titrant and the analyte.
* Direct Titration
* Indirect Titration or Replacement Method
* Back Titration
In this article, let’s focus only in the Indirect Titration or what others call Replacement Method. In Indirect Titration, the substance is not directly titrated but precipitated or removed from the reaction and the product formed has to be titrated. It uses a preliminary reaction wherein the analyte reacts with a reagent (added in large excess and need not be accurately measured); the product of this preliminary reaction is the one that reacts with the titrant.
Preliminary Reaction:
aA + rR pP
Titration Reaction:
pP + tT fF
Example:
A 0.2711 g sample of primary standard grade HgO was dissolved in an aqueous solution of KI. The liberated OH needed 13.22 mL of HCl to reach the end point. Calculate the molarity of HCl solution.
Preliminary Reaction: HgO + 4I- + H2O HgI4-2 + 2OH-
Titration Reaction: OH + H+ H2O
1 mol HgO 2 mol OH-
1 mol OH- 1 mol H+
To calculate the Molarity of HCl solution, first, we need to solve for the amount of mmol of it.
0.2711 g HgO x 1 mol HgO x 4 mol KI x 2 mol OH- x 1 mol HCl x 1000 mmol HCl
216.59 g HgO 1 mol HgO 4 mol KI 1 mol OH- 1 mol HCl
= 2.50 mmol HCl
And then we can now calculate the molarity of the HCl solution by dividing the mmol of HCl to the mL of HCl.
MHCl = mmol HCl
13.22 mL HCl
= 2.50 mmol HCl
13.22 ml HCl
= 0.19 M of HCl or 0.19 mmol/mL of HCl
By: Rhea Ciarina B. Ramos
Direct Redox Titration
Redox titrations are based on a reduction-oxidation reaction between an oxidizing agent and a reducing agent. A potentiometer or a redox indicator is usually used to determine the endpoint of the titration, as when one of the constituents is the oxidizing agent potassium dichromate. The color change of the solution from orange to green is not definite, therefore an indicator such as sodium diphenylamine is used. Analysis of wines for sulfur dioxide requires iodine as an oxidizing agent. In this case, starch is used as an indicator; a blue starch-iodine complex is formed in the presence of excess iodine, signalling the endpoint.
Some redox titrations do not require an indicator, due to the intense color of the constituents. For instance, in permanganometry a slight faint persisting pink color signals the endpoint of the titration because of the color of the excess oxidizing agent potassium permanganate.
Here is an example of Direct Redox Titration.
Half Reaction:
( 𝐹𝑒+2 𝐹𝑒+3 + 1𝑒− ) 5
( 8𝐻+ + 5𝑒− + Mn𝑂4 𝑀𝑛+2 + 4 𝐻2O ) 1
8𝐻+ + Mn𝑂4 + 5𝐹𝑒+2 𝑀𝑛+2 + 5𝐹𝑒+3 + 4 𝐻2O
Stoichiometric Ratio:
1 mol KMn𝑂4 = 5 mol Fe
1 mol Mn𝑂4 = 5 mol 𝐹𝑒+2
1 mmol Mn𝑂4 = 5 mmol 𝐹𝑒+2
You will change the grams into mol of KMn𝑂4.
0.5585 g Fe mol of KMn𝑂4
0.5585 g Fe x 1 𝑚𝑜𝑙 𝐹𝑒55.85 𝑔 𝐹𝑒 x 1 𝑚𝑜𝑙 KMn𝑂45 𝑚𝑜𝑙 𝐹𝑒 = 2𝑥10−3 𝑚𝑜𝑙 KMn𝑂4
Then you can change it to mmoles.
2𝑥10−3 𝑚𝑜𝑙 KMn𝑂4 x 1000 𝑚𝑚𝑜𝑙1 𝑚𝑜𝑙 = 2 mmol KMn𝑂4
You will now get the Molarity by:
M = 𝑚𝑜𝑙 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝐿 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 2 𝑚𝑚𝑜𝑙33.00 𝑚𝑙
M = 0.06 M
That’s all.
By: Abegail Anne G. Reyes
Some redox titrations do not require an indicator, due to the intense color of the constituents. For instance, in permanganometry a slight faint persisting pink color signals the endpoint of the titration because of the color of the excess oxidizing agent potassium permanganate.
Here is an example of Direct Redox Titration.
Half Reaction:
( 𝐹𝑒+2 𝐹𝑒+3 + 1𝑒− ) 5
( 8𝐻+ + 5𝑒− + Mn𝑂4 𝑀𝑛+2 + 4 𝐻2O ) 1
8𝐻+ + Mn𝑂4 + 5𝐹𝑒+2 𝑀𝑛+2 + 5𝐹𝑒+3 + 4 𝐻2O
Stoichiometric Ratio:
1 mol KMn𝑂4 = 5 mol Fe
1 mol Mn𝑂4 = 5 mol 𝐹𝑒+2
1 mmol Mn𝑂4 = 5 mmol 𝐹𝑒+2
You will change the grams into mol of KMn𝑂4.
0.5585 g Fe mol of KMn𝑂4
0.5585 g Fe x 1 𝑚𝑜𝑙 𝐹𝑒55.85 𝑔 𝐹𝑒 x 1 𝑚𝑜𝑙 KMn𝑂45 𝑚𝑜𝑙 𝐹𝑒 = 2𝑥10−3 𝑚𝑜𝑙 KMn𝑂4
Then you can change it to mmoles.
2𝑥10−3 𝑚𝑜𝑙 KMn𝑂4 x 1000 𝑚𝑚𝑜𝑙1 𝑚𝑜𝑙 = 2 mmol KMn𝑂4
You will now get the Molarity by:
M = 𝑚𝑜𝑙 𝑜𝑓 𝑎𝑛𝑎𝑙𝑦𝑡𝑒𝐿 𝑜𝑓 𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
= 2 𝑚𝑚𝑜𝑙33.00 𝑚𝑙
M = 0.06 M
That’s all.
By: Abegail Anne G. Reyes
Indications for Solutions
We use many products in our daily lives. Some examples of these products are for personal hygiene, food and home maintenance. All of the products we use are either acidic or basic, therefor, each have their own specific pH value.
But, pray why must we even bother ourselves with the strain of knowing the pH value of a product? A really simple answer: because the pH value is very important in determining whether or not a substance is acidic, alkaline or neutral. Also, it helps you determine how dangerous said substance is! And what better way to know than to use pH indicators.
A pH indicator is actually a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.
With that in mind let’s discuss how an indicator can be used? Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.
Tabulated below are several common laboratory pH indicators. Indicators usually exhibit intermediate colors at pH values inside the listed transition range. For example, phenol red exhibits an orange color between pH 6.8 and pH 8.4. The transition range may shift slightly depending on the concentration of the indicator in the solution and on the temperature at which it is used.
Indicator Low pH color Transition pH range High pH color
Gentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet
Leucomalachite green (first transition) yellow 0.0–2.0 green
Leucomalachite green (second transition) green 11.6–14 colorless
Thymol blue (first transition) red 1.2–2.8 yellow
Thymol blue (second transition) yellow 8.0–9.6 blue
Methyl yellow red 2.9–4.0 yellow
Bromophenol blue yellow 3.0–4.6 purple
Congo red blue-violet 3.0–5.0 red
Methyl orange red 3.1–4.4 yellow
Screened methyl orange (first transition) red 0.0–3.2 grey
Screened methyl orange (second transition) grey 3.2–4.2 green
Bromocresol green yellow 3.8–5.4 blue
Methyl red red 4.4–6.2 yellow
Azolitmin red 4.5–8.3 blue
Bromocresol purple yellow 5.2–6.8 purple
Bromothymol blue yellow 6.0–7.6 blue
Phenol red yellow 6.4–8.0 red
Neutral red red 6.8–8.0 yellow
Indicator Low pH color Transition pH range High pH color
Naphtholphthalein colorless to reddish 7.3–8.7 greenish to blue
Cresol Red yellow 7.2–8.8 reddish-purple
Phenolphthalein colorless 8.3–10.0 fuchsia
Thymolphthalein colorless 9.3–10.5 blue
Alizarine Yellow R yellow 10.2–12.0 red
pH indicators are also bountifully found in nature. Many plants or plant parts contain chemicals from the naturally-colored anthocyanin family of compounds. They are red in acidic solutions and blue in basic. Anthocyanin can be extracted with water or other solvents from a multitude of colored plants or plant parts, including from leaves (red cabbage); flowers (geranium, poppy, or rose petals); berries (blueberries, blackcurrant); and stems (rhubarb). Extracting anthocyanin from household plants, especially red cabbage, to form a crude pH indicator is a popular introductory chemistry demonstration.
Litmus, used by alchemists in the Middle Ages and still readily available, is a naturally occurring pH indicator made from a mixture of lichen species, particularly Roccella tinctoria. The word litmus is literally from 'colored moss' in Old Norse. The color changes between red in acid solutions and blue in alkalis. The term 'litmus test' has become a widely used metaphor for any test that purports to distinguish authoritatively between alternatives.
Hydrangea macrophylla flowers can change color depending on soil acidity. In acid soils, chemical reactions occur in the soil that makes aluminium available to these plants, turning the flowers blue. In alkaline soils, these reactions cannot occur and therefore aluminium is not taken up by the plant. As a result, the flowers remain pink.
By: Brea Jaye N. Uy
But, pray why must we even bother ourselves with the strain of knowing the pH value of a product? A really simple answer: because the pH value is very important in determining whether or not a substance is acidic, alkaline or neutral. Also, it helps you determine how dangerous said substance is! And what better way to know than to use pH indicators.
A pH indicator is actually a weak acid or a weak base. The undissociated form of the indicator is a different color than the iogenic form of the indicator. An Indicator does not change color from pure acid to pure alkaline at specific hydrogen ion concentration, but rather, color change occurs over a range of hydrogen ion concentrations. This range is termed the color change interval. It is expressed as a pH range.
With that in mind let’s discuss how an indicator can be used? Weak acids are titrated in the presence of indicators which change under slightly alkaline conditions. Weak bases should be titrated in the presence of indicators which change under slightly acidic conditions.
Tabulated below are several common laboratory pH indicators. Indicators usually exhibit intermediate colors at pH values inside the listed transition range. For example, phenol red exhibits an orange color between pH 6.8 and pH 8.4. The transition range may shift slightly depending on the concentration of the indicator in the solution and on the temperature at which it is used.
Indicator Low pH color Transition pH range High pH color
Gentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet
Leucomalachite green (first transition) yellow 0.0–2.0 green
Leucomalachite green (second transition) green 11.6–14 colorless
Thymol blue (first transition) red 1.2–2.8 yellow
Thymol blue (second transition) yellow 8.0–9.6 blue
Methyl yellow red 2.9–4.0 yellow
Bromophenol blue yellow 3.0–4.6 purple
Congo red blue-violet 3.0–5.0 red
Methyl orange red 3.1–4.4 yellow
Screened methyl orange (first transition) red 0.0–3.2 grey
Screened methyl orange (second transition) grey 3.2–4.2 green
Bromocresol green yellow 3.8–5.4 blue
Methyl red red 4.4–6.2 yellow
Azolitmin red 4.5–8.3 blue
Bromocresol purple yellow 5.2–6.8 purple
Bromothymol blue yellow 6.0–7.6 blue
Phenol red yellow 6.4–8.0 red
Neutral red red 6.8–8.0 yellow
Indicator Low pH color Transition pH range High pH color
Naphtholphthalein colorless to reddish 7.3–8.7 greenish to blue
Cresol Red yellow 7.2–8.8 reddish-purple
Phenolphthalein colorless 8.3–10.0 fuchsia
Thymolphthalein colorless 9.3–10.5 blue
Alizarine Yellow R yellow 10.2–12.0 red
pH indicators are also bountifully found in nature. Many plants or plant parts contain chemicals from the naturally-colored anthocyanin family of compounds. They are red in acidic solutions and blue in basic. Anthocyanin can be extracted with water or other solvents from a multitude of colored plants or plant parts, including from leaves (red cabbage); flowers (geranium, poppy, or rose petals); berries (blueberries, blackcurrant); and stems (rhubarb). Extracting anthocyanin from household plants, especially red cabbage, to form a crude pH indicator is a popular introductory chemistry demonstration.
Litmus, used by alchemists in the Middle Ages and still readily available, is a naturally occurring pH indicator made from a mixture of lichen species, particularly Roccella tinctoria. The word litmus is literally from 'colored moss' in Old Norse. The color changes between red in acid solutions and blue in alkalis. The term 'litmus test' has become a widely used metaphor for any test that purports to distinguish authoritatively between alternatives.
Hydrangea macrophylla flowers can change color depending on soil acidity. In acid soils, chemical reactions occur in the soil that makes aluminium available to these plants, turning the flowers blue. In alkaline soils, these reactions cannot occur and therefore aluminium is not taken up by the plant. As a result, the flowers remain pink.
By: Brea Jaye N. Uy
Standardization of Acid and Base Solutions
Studying Chemistry is fun and exciting especially when it comes to experiments like mixing different chemicals to know what will their product be. But as always everything is not too simple, you need to carefully analyze concepts to fully understand Chemistry. One topic in Chemistry that needed to be familiarized with, is the Standardization of Acid and Base Solutions.
Standardization is the process of determining the exact concentration or the Molarity of a solution or it is simply the process whereby the concentration of a reagent is determined by reaction with a known quantity of a second reagent.
In this process, one type of analytical procedure is often used which is the Titration. In a titration, an exact volume of one substance is reacted with a known amount of another substance. The point at which the reaction is complete in a titration is referred to as the endpoint. A chemical substance known as an indicator is used to indicate (signal) the endpoint.
To clearly understand standardization here’s an example, an experiment involves two separate acid-base standardization procedures. In the first standardization the molarity of a sodium hydroxide solution (NaOH) will be determined by titrating a sample of potassium acid phthalate (KHP; HKC8H4O4) with the NaOH. In the second procedure the standardized NaOH will be used to determine the molarity of a hydrochloric solution (HCl).
The indicator used in this experiment is phenolphthalein. Phenolphthalein, an organic compound, is colorless in acidic solution and pink in basic solution.
For the first standardization, a 0.128 g sample of KHP (HKC8H4O4) required 28.54 mL of NaOH solution to reach a phenolphthalein endpoint. You need to calculate the molarity of the NaOH using the balanced chemical equation.
HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O
First, convert the grams of KHP to its molarity by multiplying its given mass to the number of moles then dividing to its formula mass to cancel same units.
(0.128 g KHP)(1 mol / 204.23 g KHP ) = 6.267 x 10-4 mol KHP
After you get the number of moles of KHP, convert it to the number of moles of NaOH by multiplying the obtained moles of KHP to the moles of NaOH then dividing it to the number of moles of KHP according to their ratio (1:1) in the balanced equation.
(6.267 x 10-4 mol KHP)(1 mol NaOH / 1 mol KHP) = 6.267 x 10-4 mol NaOH
Lastly, to get the molarity of NaOH, divide the obtained number of moles of NaOH to the given volume of NaOH.
6.267 x 10-4 mol NaOH / 0.02854 L NaOH = 0.0220 M NaOH
For the second standardization, a 20.00 mL sample of HCl was titrated with the NaOH solution from first standardization. To reach the endpoint required 23.72 mL of the NaOH. Again you need to calculate the molarity of the HCl using their balanced chemical equation.
HCl + NaOH -----> NaCl + H2O
First, convert the given volume from mL to L by dividing to 1000. Then multiply the given volume in L of NaOH to the obtained molarity of NaOH from the first standardization. Next, divide the obtained molarity to the number of L of NaOH to cancel same units.
(0.02372 L NaOH)(0.0220 mol NaOH / 1 L NaOH) = 5.218 x 10-4 mol NaOH
Then to convert moles of NaOH to moles of HCl, multiply the obtained number of moles of NaOH to the number of moles of HCl then divide it to the moles of NaOH according to the ratio (1:1) in their balanced equation.
(5.218 x 10-4 mol NaOH)(1 mol HCl / 1 mol NaOH) = 5.218 x 10-4 mol HCl
Lastly, to get the Molarity of HCl, divide the obtained number of moles of HCl to its given volume.
5.218 x 10-4 mol HCl / 0.02000 L HCl = 0.0261 M HCl
That ‘s how to perform the calculations in Standardization of Acid and Base solutions.
References: http://www.chem.latech.edu http://en.wikipedia.org
By: Joanna M. Mayo
Standardization is the process of determining the exact concentration or the Molarity of a solution or it is simply the process whereby the concentration of a reagent is determined by reaction with a known quantity of a second reagent.
In this process, one type of analytical procedure is often used which is the Titration. In a titration, an exact volume of one substance is reacted with a known amount of another substance. The point at which the reaction is complete in a titration is referred to as the endpoint. A chemical substance known as an indicator is used to indicate (signal) the endpoint.
To clearly understand standardization here’s an example, an experiment involves two separate acid-base standardization procedures. In the first standardization the molarity of a sodium hydroxide solution (NaOH) will be determined by titrating a sample of potassium acid phthalate (KHP; HKC8H4O4) with the NaOH. In the second procedure the standardized NaOH will be used to determine the molarity of a hydrochloric solution (HCl).
The indicator used in this experiment is phenolphthalein. Phenolphthalein, an organic compound, is colorless in acidic solution and pink in basic solution.
For the first standardization, a 0.128 g sample of KHP (HKC8H4O4) required 28.54 mL of NaOH solution to reach a phenolphthalein endpoint. You need to calculate the molarity of the NaOH using the balanced chemical equation.
HKC8H4O4 + NaOH -----> NaKC8H4O4 + H2O
First, convert the grams of KHP to its molarity by multiplying its given mass to the number of moles then dividing to its formula mass to cancel same units.
(0.128 g KHP)(1 mol / 204.23 g KHP ) = 6.267 x 10-4 mol KHP
After you get the number of moles of KHP, convert it to the number of moles of NaOH by multiplying the obtained moles of KHP to the moles of NaOH then dividing it to the number of moles of KHP according to their ratio (1:1) in the balanced equation.
(6.267 x 10-4 mol KHP)(1 mol NaOH / 1 mol KHP) = 6.267 x 10-4 mol NaOH
Lastly, to get the molarity of NaOH, divide the obtained number of moles of NaOH to the given volume of NaOH.
6.267 x 10-4 mol NaOH / 0.02854 L NaOH = 0.0220 M NaOH
For the second standardization, a 20.00 mL sample of HCl was titrated with the NaOH solution from first standardization. To reach the endpoint required 23.72 mL of the NaOH. Again you need to calculate the molarity of the HCl using their balanced chemical equation.
HCl + NaOH -----> NaCl + H2O
First, convert the given volume from mL to L by dividing to 1000. Then multiply the given volume in L of NaOH to the obtained molarity of NaOH from the first standardization. Next, divide the obtained molarity to the number of L of NaOH to cancel same units.
(0.02372 L NaOH)(0.0220 mol NaOH / 1 L NaOH) = 5.218 x 10-4 mol NaOH
Then to convert moles of NaOH to moles of HCl, multiply the obtained number of moles of NaOH to the number of moles of HCl then divide it to the moles of NaOH according to the ratio (1:1) in their balanced equation.
(5.218 x 10-4 mol NaOH)(1 mol HCl / 1 mol NaOH) = 5.218 x 10-4 mol HCl
Lastly, to get the Molarity of HCl, divide the obtained number of moles of HCl to its given volume.
5.218 x 10-4 mol HCl / 0.02000 L HCl = 0.0261 M HCl
That ‘s how to perform the calculations in Standardization of Acid and Base solutions.
References: http://www.chem.latech.edu http://en.wikipedia.org
By: Joanna M. Mayo
Titration
We all know that there are different types of titration. But before we proceed with that, let’s just first define what is titration. So, do you have any idea when you heard the word “titration”?
In chemistry, Titration, also known as titrimetry, is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Because volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand to determine concentration. It is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction.
There are different types of titration namely acid-base titration, and redox titration. Let us discuss them one by one.
In acid-base titration, is the determination of the concentration of an acid or base by exactly neutralizing the acid or base with an acid or base of known concentration. This allows for quantitative analysis of the concentration of an unknown acid or base solution. It makes use of the neutralization reaction that occurs between acids and bases and the knowledge of how acids and bases will react if their formulas are known. Before starting the titration a suitable pH indicator must be chosen. The equivalence point of the reaction, the point at which equivalent amounts of the reactants have reacted, will have a pH dependent on the relative strengths of the acid and base used. The pH of the equivalence point can be estimated using the following rules:
* A strong acid will react with a strong base to form a neutral (pH = 7) solution.
* A strong acid will react with a weak base to form an acidic (pH < 7) solution.
* A weak acid will react with a strong base to form a basic (pH > 7) solution.
Redox titration (also called oxidation-reduction titration) is a type of titration based on a redox reaction between the analyte and titrant. Redox titration may involve the use of a redox indicator and/or a potentiometer.
REFERENCES:
http://en.wikipedia.org/wiki/Titration http://www.dartmouth.edu/~chemlab/techniques/titration.html http://en.wikipedia.org/wiki/Acid%E2%80%93base_titration http://en.wikipedia.org/wiki/Redox_titration
By: Dannah Lyn R. Antonio
In chemistry, Titration, also known as titrimetry, is a common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte. Because volume measurements play a key role in titration, it is also known as volumetric analysis. A reagent, called the titrant or titrator is prepared as a standard solution. A known concentration and volume of titrant reacts with a solution of analyte or titrand to determine concentration. It is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction.
There are different types of titration namely acid-base titration, and redox titration. Let us discuss them one by one.
In acid-base titration, is the determination of the concentration of an acid or base by exactly neutralizing the acid or base with an acid or base of known concentration. This allows for quantitative analysis of the concentration of an unknown acid or base solution. It makes use of the neutralization reaction that occurs between acids and bases and the knowledge of how acids and bases will react if their formulas are known. Before starting the titration a suitable pH indicator must be chosen. The equivalence point of the reaction, the point at which equivalent amounts of the reactants have reacted, will have a pH dependent on the relative strengths of the acid and base used. The pH of the equivalence point can be estimated using the following rules:
* A strong acid will react with a strong base to form a neutral (pH = 7) solution.
* A strong acid will react with a weak base to form an acidic (pH < 7) solution.
* A weak acid will react with a strong base to form a basic (pH > 7) solution.
Redox titration (also called oxidation-reduction titration) is a type of titration based on a redox reaction between the analyte and titrant. Redox titration may involve the use of a redox indicator and/or a potentiometer.
REFERENCES:
http://en.wikipedia.org/wiki/Titration http://www.dartmouth.edu/~chemlab/techniques/titration.html http://en.wikipedia.org/wiki/Acid%E2%80%93base_titration http://en.wikipedia.org/wiki/Redox_titration
By: Dannah Lyn R. Antonio
Back Titration
A back titration, or indirect titration, is generally a two-stage analytical technique:
a. Reactant A of unknown concentration is reacted with excess reactant B of known concentration. b. A titration is then performed to determine the amount of reactant B in excess.
Back titrations are used when:
* one of the reactants is volatile, for example ammonia.
* an acid or a base is an insoluble salt, for example calcium carbonate
* a particular reaction is too slow * direct titration would involve a weak acid - weak base titration (the end-point of this type of direct titration is very difficult to observe)
Example : Back (Indirect) Titration to Determine the Concentration of a Volatile Substance
A student was asked to determine the concentration of ammonia, a volatile substance, in a commercially available cloudy ammonia solution used for cleaning. First the student pipetted 25.00mL of the cloudy ammonia solution into a 250.0mL conical flask. 50.00mL of 0.100M HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution. The excess (unreacted) HCl was then titrated with 0.050M Na2CO3(aq). 21.50mL of Na2CO3(aq) was required. Calculate the concentration of the ammonia in the cloudy ammonia solution.
Step 1: Determine the amount of HCl in excess from the titration results
a. Write the equation for the titration:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
acid + carbonate → salt + carbon dioxide + water
b. Calculate the moles, n, of Na2CO3(aq) that reacted in the titration: n = M x V M = 0.050 molL-1 V = 21.50mL = 21.50 x 10-3L n(Na2CO3(aq)) = 0.050 x 21.50 x 10-3 = 1.075 x 10-3 mol
c. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration. From the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl So, 1.075 x 10-3 mole Na2CO3 reacted with 2 x 1.075 x 10-3 moles HCl n(HCltitrated) = 2 x 1.075 x 10-3 = 2.150 x 10-3 mol
d. The amount of HCl that was added to the cloudy ammonia solution in excess was 2.150 x 10-3 mol
Step 2: Determine the amount of ammonia in the cloudy ammonia solution
a. Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution: n(HCltotal added) = M x V M = 0.100 molL-1
V = 50.00mL = 50.00 x 10-3L n(HCltotal added) = 0.100 x 50.00 x 10-3 = 5.00 x 10-3 mol
b. Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution n(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added) n(HCltotal added) = 5.00 x 10-3 mol n(HCltitrated) = 2.150 x 10-3 mol 2.150 x 10-3 + n(HClreacted with ammonia) = 5.00 x 10-3 n(HClreacted with ammonia) = 5.00 x 10-3 - 2.150 x 10-3 = 2.85 x 10-3 mol
c. Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).
NH3(aq) + HCl(aq) → NH4Cl(aq)
d. From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl. From the equation, 1 mol HCl reacts with 1 mol NH3 So, 2.85 x 10-3 mol HCl had reacted with 2.85 x 10-3 mol NH3 in the cloudy ammonia solution.
e. Calculate the ammonia concentration in the cloudy ammonia solution. M = n ÷ V n = 2.85 x 10-3 mol (moles of NH3 that reacted with HCl) V = 25.00mL = 25.00 x 10-3L (volume of ammonia solution that reacted with HCl) M = 2.85 x 10-3 ÷ 25.00 x 10-3 = 0.114 M
f. The concentration of ammonia in the cloudy ammonia solution was 0.114M
Example : Back (Indirect) Titration to Determine the Amount of an Insoluble Salt
A student was asked to determine the mass, in grams, of calcium carbonate present in a 0.125g sample of chalk. The student placed the chalk sample in a 250mL conical flask and added 50.00mL 0.200M HCl using a pipette. The excess HCl was then titrated with 0.250M NaOH. The average NaOH titre was 32.12mL Calculate the mass of calcium carbonate, in grams, present in the chalk sample.
Step 1: Determine the amount of HCl in excess from the titration results
a. Write the equation for the titration:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
b. Calculate the moles, n, of NaOH(aq) that reacted in the titration: n = M x V M = 0.250 molL-1 V = 32.12mL = 32.12 x 10-3L n(NaOH(aq)) = 0.250 x 32.12 x 10-3 = 8.03 x 10-3 mol
c. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration. From the balanced chemical equation, 1 mole NaOH reacts with 1 mole of HCl So, 8.03 x 10-3 mole NaOH reacted with 8.03 x 10-3 moles HCl
d. The amount of HCl that was added to the chalk in excess was 8.03 x 10-3 mol
Step 2: Determine the amount of calcium carbonate in chalk
a. Calculate the total moles of HCl originally added to the chalk: n(HCltotal added) = M x V M = 0.200 molL-1 V = 50.00mL = 50.00 x 10-3L n(HCltotal added) = 0.200 x 50.00 x 10-3 = 0.010 mol
b. Calculate the moles of HCl that reacted with the calcium carbonate in the chalk n(HCltitrated) + n(HClreacted with calcium carbonate) = n(HCltotal added) n(HCltotal added) = 0.010 mol n(HCltitrated) = 8.03 x 10-3 mol 8.03 x 10-3 + n(HClreacted with calcium carbonate) = 0.010 n(HClreacted with calcium carbonate) = 0.010 - 8.03 x 10-3 = 1.97 x 10-3 mol
c. Write the balanced chemical equation for the reaction between calcium carbonate in the chalk and the HCl(aq).
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
d. From the balanced chemical equation, calculate the moles of CaCO3 that reacted with HCl. From the equation, 1 mol CaCO3 reacts with 2 mol HCl so, 1 mol HCl reacts with ½ mol CaCO3 So, 1.97 x 10-3 mol HCl had reacted with ½ x 1.97 x 10-3 = 9.85 x 10-4 mol CaCO3 in the chalk.
e. Calculate the mass of calcium carbonate in the chalk. n = mass ÷ MM n = 9.85 x 10-4 mol (moles of CaCO3 that reacted with HCl) MM(CaCO3) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol mass = n x MM = 9.85 x 10-4 x 100.09 = 0.099g
f. The mass of calcium carbonate in the chalk was 0.099g
By: Jillian Carlo N. Gervacio
a. Reactant A of unknown concentration is reacted with excess reactant B of known concentration. b. A titration is then performed to determine the amount of reactant B in excess.
Back titrations are used when:
* one of the reactants is volatile, for example ammonia.
* an acid or a base is an insoluble salt, for example calcium carbonate
* a particular reaction is too slow * direct titration would involve a weak acid - weak base titration (the end-point of this type of direct titration is very difficult to observe)
Example : Back (Indirect) Titration to Determine the Concentration of a Volatile Substance
A student was asked to determine the concentration of ammonia, a volatile substance, in a commercially available cloudy ammonia solution used for cleaning. First the student pipetted 25.00mL of the cloudy ammonia solution into a 250.0mL conical flask. 50.00mL of 0.100M HCl(aq) was immediately added to the conical flask which reacted with the ammonia in solution. The excess (unreacted) HCl was then titrated with 0.050M Na2CO3(aq). 21.50mL of Na2CO3(aq) was required. Calculate the concentration of the ammonia in the cloudy ammonia solution.
Step 1: Determine the amount of HCl in excess from the titration results
a. Write the equation for the titration:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
acid + carbonate → salt + carbon dioxide + water
b. Calculate the moles, n, of Na2CO3(aq) that reacted in the titration: n = M x V M = 0.050 molL-1 V = 21.50mL = 21.50 x 10-3L n(Na2CO3(aq)) = 0.050 x 21.50 x 10-3 = 1.075 x 10-3 mol
c. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration. From the balanced chemical equation, 1 mole Na2CO3 react with 2 moles of HCl So, 1.075 x 10-3 mole Na2CO3 reacted with 2 x 1.075 x 10-3 moles HCl n(HCltitrated) = 2 x 1.075 x 10-3 = 2.150 x 10-3 mol
d. The amount of HCl that was added to the cloudy ammonia solution in excess was 2.150 x 10-3 mol
Step 2: Determine the amount of ammonia in the cloudy ammonia solution
a. Calculate the total moles of HCl originally added to the diluted cloudy ammonia solution: n(HCltotal added) = M x V M = 0.100 molL-1
V = 50.00mL = 50.00 x 10-3L n(HCltotal added) = 0.100 x 50.00 x 10-3 = 5.00 x 10-3 mol
b. Calculate the moles of HCl that reacted with the ammonia in the diluted cloudy ammonia solution n(HCltitrated) + n(HClreacted with ammonia) = n(HCltotal added) n(HCltotal added) = 5.00 x 10-3 mol n(HCltitrated) = 2.150 x 10-3 mol 2.150 x 10-3 + n(HClreacted with ammonia) = 5.00 x 10-3 n(HClreacted with ammonia) = 5.00 x 10-3 - 2.150 x 10-3 = 2.85 x 10-3 mol
c. Write the balanced chemical equation for the reaction between ammonia in the cloudy ammonia solution and the HCl(aq).
NH3(aq) + HCl(aq) → NH4Cl(aq)
d. From the balanced chemical equation, calculate the moles of NH3 that reacted with HCl. From the equation, 1 mol HCl reacts with 1 mol NH3 So, 2.85 x 10-3 mol HCl had reacted with 2.85 x 10-3 mol NH3 in the cloudy ammonia solution.
e. Calculate the ammonia concentration in the cloudy ammonia solution. M = n ÷ V n = 2.85 x 10-3 mol (moles of NH3 that reacted with HCl) V = 25.00mL = 25.00 x 10-3L (volume of ammonia solution that reacted with HCl) M = 2.85 x 10-3 ÷ 25.00 x 10-3 = 0.114 M
f. The concentration of ammonia in the cloudy ammonia solution was 0.114M
Example : Back (Indirect) Titration to Determine the Amount of an Insoluble Salt
A student was asked to determine the mass, in grams, of calcium carbonate present in a 0.125g sample of chalk. The student placed the chalk sample in a 250mL conical flask and added 50.00mL 0.200M HCl using a pipette. The excess HCl was then titrated with 0.250M NaOH. The average NaOH titre was 32.12mL Calculate the mass of calcium carbonate, in grams, present in the chalk sample.
Step 1: Determine the amount of HCl in excess from the titration results
a. Write the equation for the titration:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
b. Calculate the moles, n, of NaOH(aq) that reacted in the titration: n = M x V M = 0.250 molL-1 V = 32.12mL = 32.12 x 10-3L n(NaOH(aq)) = 0.250 x 32.12 x 10-3 = 8.03 x 10-3 mol
c. Use the balanced chemical reaction for the titration to determine the moles of HCl that reacted in the titration. From the balanced chemical equation, 1 mole NaOH reacts with 1 mole of HCl So, 8.03 x 10-3 mole NaOH reacted with 8.03 x 10-3 moles HCl
d. The amount of HCl that was added to the chalk in excess was 8.03 x 10-3 mol
Step 2: Determine the amount of calcium carbonate in chalk
a. Calculate the total moles of HCl originally added to the chalk: n(HCltotal added) = M x V M = 0.200 molL-1 V = 50.00mL = 50.00 x 10-3L n(HCltotal added) = 0.200 x 50.00 x 10-3 = 0.010 mol
b. Calculate the moles of HCl that reacted with the calcium carbonate in the chalk n(HCltitrated) + n(HClreacted with calcium carbonate) = n(HCltotal added) n(HCltotal added) = 0.010 mol n(HCltitrated) = 8.03 x 10-3 mol 8.03 x 10-3 + n(HClreacted with calcium carbonate) = 0.010 n(HClreacted with calcium carbonate) = 0.010 - 8.03 x 10-3 = 1.97 x 10-3 mol
c. Write the balanced chemical equation for the reaction between calcium carbonate in the chalk and the HCl(aq).
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
d. From the balanced chemical equation, calculate the moles of CaCO3 that reacted with HCl. From the equation, 1 mol CaCO3 reacts with 2 mol HCl so, 1 mol HCl reacts with ½ mol CaCO3 So, 1.97 x 10-3 mol HCl had reacted with ½ x 1.97 x 10-3 = 9.85 x 10-4 mol CaCO3 in the chalk.
e. Calculate the mass of calcium carbonate in the chalk. n = mass ÷ MM n = 9.85 x 10-4 mol (moles of CaCO3 that reacted with HCl) MM(CaCO3) = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol mass = n x MM = 9.85 x 10-4 x 100.09 = 0.099g
f. The mass of calcium carbonate in the chalk was 0.099g
By: Jillian Carlo N. Gervacio
Titrations
Titration, also known as titrimetry, A titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH Meter is used to detect the endpoint of the reaction.
HOW TO TITRATE ?
Doing a Titration
* Begin by preparing your buret, as described on theburet page. Your buret should be conditioned and filled with titrant solution. You should check for air bubbles and leaks, before proceding with the titration.
Take an initial volume reading and record it in your notebook. Before beginning a titration, you should always calculate the expected endpoint volume.
Prepare the solution to be analyzed by placing it in a clean Erlenmeyer flask or beaker. If your sample is a solid, make sure it is completely dissoloved. Put a magnetic stirrer in the flask and add indicator.
Use the buret to deliver a stream of titrant to within a couple of mL of your expected endpoint. You will see the indicator change color when the titrant hits the solution in the flask, but the color change disappears upon stirring.
Approach the endpoint more slowly and watch the color of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the buret, to be sure all titrant is mixed in the flask.
As you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.
Make sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink. The pale pink fades in 10 to 20 minutes. If you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint.
If the flask looks like this, you have gone too far!
When you have reached the endpoint, read the final volume in the buret and record it in your notebook.
Subtract the initial volume to determine the amount of titrant delivered. Use this, the concentration of the titrant, and the stoichiometry of the titration reaction to calculate the number of moles of reactant in your analyte solution.
Titrating with a pH meter
* Titration with a pH meter follows the same procedure as a titration with an indicator, except that the endpoint is detected by a rapid change in pH, rather than the color change of an indicator.
Arrange the sample, stirrer, buret, and pH meter electrode so that you can read the pH and operate the buret with ease.
To detect the endpoint accurately, record pH vs. volume of titrant added and plot the titration curve as you titrate.
By: Princes Joy A. Nunez
HOW TO TITRATE ?
Doing a Titration
* Begin by preparing your buret, as described on theburet page. Your buret should be conditioned and filled with titrant solution. You should check for air bubbles and leaks, before proceding with the titration.
Take an initial volume reading and record it in your notebook. Before beginning a titration, you should always calculate the expected endpoint volume.
Prepare the solution to be analyzed by placing it in a clean Erlenmeyer flask or beaker. If your sample is a solid, make sure it is completely dissoloved. Put a magnetic stirrer in the flask and add indicator.
Use the buret to deliver a stream of titrant to within a couple of mL of your expected endpoint. You will see the indicator change color when the titrant hits the solution in the flask, but the color change disappears upon stirring.
Approach the endpoint more slowly and watch the color of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the buret, to be sure all titrant is mixed in the flask.
As you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.
Make sure you know what the endpoint should look like. For phenolphthalein, the endpoint is the first permanent pale pink. The pale pink fades in 10 to 20 minutes. If you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint.
If the flask looks like this, you have gone too far!
When you have reached the endpoint, read the final volume in the buret and record it in your notebook.
Subtract the initial volume to determine the amount of titrant delivered. Use this, the concentration of the titrant, and the stoichiometry of the titration reaction to calculate the number of moles of reactant in your analyte solution.
Titrating with a pH meter
* Titration with a pH meter follows the same procedure as a titration with an indicator, except that the endpoint is detected by a rapid change in pH, rather than the color change of an indicator.
Arrange the sample, stirrer, buret, and pH meter electrode so that you can read the pH and operate the buret with ease.
To detect the endpoint accurately, record pH vs. volume of titrant added and plot the titration curve as you titrate.
By: Princes Joy A. Nunez
Titrations
A titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the "analyte") until the equivalence point is reached. The equivalence point is the point at which titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed. Otherwise, an indicator may be added which has an "endpoint" (changes color) at the equivalence point, or the equivalence point may be determined from a titration curve.
The amount of added titrant is determined from its concentration and volume:
n(mol) = C(mol/L) x V(L)
and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte. A measured volume of the solution to be titrated, in this case, colorless aqueous acetic acid, CH3COOH (aq) is placed in a beaker. The colorless sodium hydroxide NaOH (aq), which is the titrant, is added carefully by means of a buret. The volume of titrant added can then be determined by reading the level of liquid in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, so precise additions of titrant can be made rapidly.
As the first few milliliters of titrant flow into the flask, some indicator briefly changes to pink, but returns to colorless rapidly. This is due to a large excess of acetic acid. Thelimiting reagent NaOH is entirely consumed.
The added indicator changes to pink when the titration is complete, indicating that all of the aqueous acetic acid has been consumed by NaOH(aq). The reaction which occurs is
CH3COOH(aq) + NaOH(aq) → Na+(aq) + CH3COO-(aq) + H2O(l) Eq (1)
Eventually, all the acetic acid is consumed. Addition of even a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The color change that occurs at the endpoint of the indicator signals that all the acetic acid has been consumed, so we have reached the equivalence point of the titration. If slightly more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The endpoint appears suddenly, and care must be taken not to overshoot the endpoint.
After the titration has reached the endpoint, a final volume is read from the buret. Using the initial and final reading, the volume added can be determined quite precisely:
The object of a titration is always to add just the amount of titrant needed to consume exactly the amount of substance being titrated. In the NaOH—CH3COOH reaction [Eq. (1)], the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio
Let’s do an example:
What volume of 0.05386 M KMnO4 would be needed to reach the endpoint when titrating 25.00 ml of 0.1272 M H2O2, given S(KMnO4/H2O2) = 2/5?
At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of KMnO4 which must be added:
The amount of H2O2 is obtained from the volume and concentration:
Then,
= 1.272mmol KMnO4
To obtain VKMnO4(aq) we use the concentration as a conversion factor:
= 23.62 cm3
Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4.
References: http://www.sciensational.com
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Titrations-875.html
By: Lorie Anne R. Anda
The amount of added titrant is determined from its concentration and volume:
n(mol) = C(mol/L) x V(L)
and the amount of titrant can be used in the usual stoichiometric calculation to determine the amount of analyte. A measured volume of the solution to be titrated, in this case, colorless aqueous acetic acid, CH3COOH (aq) is placed in a beaker. The colorless sodium hydroxide NaOH (aq), which is the titrant, is added carefully by means of a buret. The volume of titrant added can then be determined by reading the level of liquid in the buret before and after titration. This reading can usually be estimated to the nearest hundredth of a milliliter, so precise additions of titrant can be made rapidly.
As the first few milliliters of titrant flow into the flask, some indicator briefly changes to pink, but returns to colorless rapidly. This is due to a large excess of acetic acid. Thelimiting reagent NaOH is entirely consumed.
The added indicator changes to pink when the titration is complete, indicating that all of the aqueous acetic acid has been consumed by NaOH(aq). The reaction which occurs is
CH3COOH(aq) + NaOH(aq) → Na+(aq) + CH3COO-(aq) + H2O(l) Eq (1)
Eventually, all the acetic acid is consumed. Addition of even a fraction of a drop of titrant produces a lasting pink color due to unreacted NaOH in the flask. The color change that occurs at the endpoint of the indicator signals that all the acetic acid has been consumed, so we have reached the equivalence point of the titration. If slightly more NaOH solution were added, there would be an excess and the color of the solution in the flask would get much darker. The endpoint appears suddenly, and care must be taken not to overshoot the endpoint.
After the titration has reached the endpoint, a final volume is read from the buret. Using the initial and final reading, the volume added can be determined quite precisely:
The object of a titration is always to add just the amount of titrant needed to consume exactly the amount of substance being titrated. In the NaOH—CH3COOH reaction [Eq. (1)], the equivalence point occurs when an equal molar amount of NaOH has been added from the graduated cylinder for every mole of CH3COOH originally in the titration flask. That is, at the equivalence point the ratio of the amount of NaOH, added to the amount of CH3COOH consumed must equal the stoichiometric ratio
Let’s do an example:
What volume of 0.05386 M KMnO4 would be needed to reach the endpoint when titrating 25.00 ml of 0.1272 M H2O2, given S(KMnO4/H2O2) = 2/5?
At the equivalence point, the stoichiometric ratio will apply, and we can use it to calculate the amount of KMnO4 which must be added:
The amount of H2O2 is obtained from the volume and concentration:
Then,
= 1.272mmol KMnO4
To obtain VKMnO4(aq) we use the concentration as a conversion factor:
= 23.62 cm3
Note that overtitrating [adding more than 23.62 cm3 of KMnO4(aq) would involve an excess (more than 1.272 mmol) of KMnO4.
References: http://www.sciensational.com
http://chemed.chem.wisc.edu/chempaths/GenChem-Textbook/Titrations-875.html
By: Lorie Anne R. Anda