AN INTRODUCTION TO TITRATION
Titration is a process of chemical analysis in which the quantity of some component of a sample is determined by adding to the measured sample an exactly known quantity of another substance with which the desired component reacts in a definite, known amount. The process is usually carried out by gradually adding a standard solution, a solution of a known or unknown substance, of titrating reagent, or titrant, from a burette, basically a long, graduated measuring tube with a faucet and a transfer tube at its lower end. The addition is stopped when the equivalence point is reached.
This so called equivalence point is the point at which an added titrant is stoichiometrically equal to the number of moles of substance, known as analyte, present in the sample: the smallest amount of titrant that is sufficient to fully neutralize or react with the analyte. At the equivalence point of a titration, an exactly equivalent amount of titrant has been added to the sample. The experimental point at which the completion of the reaction is marked by some signal is called the end point. This signal can be the colour change of an indicator or a change in some property that is measured throughout the titration. The difference between the end point and the equivalence point is the titration error, which is kept as small as possible by the appropriate choice of an end-point signal and a method for distinguishing it.
For many titration reactions it is possible to find a suitable visual colour indicator that will signal the end point at, or very close to, the equivalence point. Such titrations, classified according to the nature of the chemical reaction occurring between the sample and titrant, include: acid-base titrations, precipitation titrations, complex-formation titrations, and oxidation-reduction (redox) titrations. In acid-base titration, the indicator is a substance that can exist in two forms, an acid form and a basic form, which differ in colour. For example, litmus is blue in alkaline solution and red in acid solution. Phenolphthalein is colourless in acid solution and red in alkaline solution. A wide choice of acid-base indicators is available, varying not only in the colours of the two forms but also in their sensitivity toward acid or base. As the analyte is being titrated its color changes.
An indicator is added to the solution being titrated. The indicator is a substance that changes color when the reaction is complete. In our example, phenolphthalein, which is a commonly used acid‑base indicator, is added to the nitric acid solution in the Erlenmeyer flask. Phenolphthalein has two chemical forms. In acidic conditions, it is in the acid form, which is colorless. In basic conditions, an H+ ion is removed from each phenolphthalein molecule, converting it to its base form, which is red.
The titrant is slowly added to the solution being titrated until the indicator changes color, showing that the reaction is complete. This stage in the procedure is called the endpoint. In our example, the NaOH solution is slowly added from the burette until the mixture in the Erlenmeyer flask changes from colorless to red. The OH− ions in the NaOH solution react with the H3O+ ions in the HNO3 solution.
H3O+(aq) + OH−(aq) → H2O(l)
As long as there are excess H3O+ ions in the solution, the solution stays acidic, the phenolphthalein stays mostly in the acid form, and the solution is colorless. When enough NaOH solution is added to react with all of the H3O+ ions, the reaction is complete. When a small amount of extra NaOH solution is added, perhaps one drop, there will be an excess of hydroxide ions, OH−, in solution. These react with the phenolphthalein molecules, changing them from the acid form to the base form. Because the base form is red, the solution turns red, telling us that the reaction is complete (or just slightly beyond complete).
The volume of titrant added from the buret is measured. For our example, let’s assume that 18.3 mL of 0.115 M NaOH has been added. The following setup shows how the molarity of the nitric acid solution can be calculated from this data.
? mol HNO3 = 18.3 ml NaOH x 1000 ml x 0.115 mol NaOH x 1 mol HNO3
1 L HNO3 25.00 ml HNO3 1L 1000 ml NaOH 1 mol NaOH
= or 0.0842 M HNO3
The first step the unit analysis thought-process is to clearly identify the units that you want. Molarity describes the number of moles of solute per liter of solution, so we start by placing moles of HNO3 over 1 L HNO3 solution.
Because molarity is a ratio of two units, we begin our calculation with a ratio of two units. Knowing that we want volume of HNO3 solution on the bottom when we are done, we place 25.00 mL HNO3 solution on the bottom at the start. We place 18.3 mL NaOH solution on the top of our ratio, giving us the ratio of two units overall that we want.
We convert milliliters of HNO3 solution to liters of HNO3 solution using the relationship between milliliters and liters. The last two conversion factors convert from amount of one substance in a chemical reaction (mL NaOH solution) to amount of another substance in the reaction (mol HNO3). Thus this is an equation stoichiometry problem that requires at its core the conversion of moles of NaOH to moles of HNO3 using the molar ratio for the reaction between them.
NaOH(aq) + HNO3(aq) NaNO3(aq) + H2O(l)
In order to use the molar ratio to convert from moles of NaOH to moles of HNO3, we need to convert from volume of NaOH solution to moles of NaOH using the molarity as a conversion factor.
This is a basic Titration problem, the fundamentals of titration.
-Leomar T. Santiago
IV-Zara, Group5
This so called equivalence point is the point at which an added titrant is stoichiometrically equal to the number of moles of substance, known as analyte, present in the sample: the smallest amount of titrant that is sufficient to fully neutralize or react with the analyte. At the equivalence point of a titration, an exactly equivalent amount of titrant has been added to the sample. The experimental point at which the completion of the reaction is marked by some signal is called the end point. This signal can be the colour change of an indicator or a change in some property that is measured throughout the titration. The difference between the end point and the equivalence point is the titration error, which is kept as small as possible by the appropriate choice of an end-point signal and a method for distinguishing it.
For many titration reactions it is possible to find a suitable visual colour indicator that will signal the end point at, or very close to, the equivalence point. Such titrations, classified according to the nature of the chemical reaction occurring between the sample and titrant, include: acid-base titrations, precipitation titrations, complex-formation titrations, and oxidation-reduction (redox) titrations. In acid-base titration, the indicator is a substance that can exist in two forms, an acid form and a basic form, which differ in colour. For example, litmus is blue in alkaline solution and red in acid solution. Phenolphthalein is colourless in acid solution and red in alkaline solution. A wide choice of acid-base indicators is available, varying not only in the colours of the two forms but also in their sensitivity toward acid or base. As the analyte is being titrated its color changes.
An indicator is added to the solution being titrated. The indicator is a substance that changes color when the reaction is complete. In our example, phenolphthalein, which is a commonly used acid‑base indicator, is added to the nitric acid solution in the Erlenmeyer flask. Phenolphthalein has two chemical forms. In acidic conditions, it is in the acid form, which is colorless. In basic conditions, an H+ ion is removed from each phenolphthalein molecule, converting it to its base form, which is red.
The titrant is slowly added to the solution being titrated until the indicator changes color, showing that the reaction is complete. This stage in the procedure is called the endpoint. In our example, the NaOH solution is slowly added from the burette until the mixture in the Erlenmeyer flask changes from colorless to red. The OH− ions in the NaOH solution react with the H3O+ ions in the HNO3 solution.
H3O+(aq) + OH−(aq) → H2O(l)
As long as there are excess H3O+ ions in the solution, the solution stays acidic, the phenolphthalein stays mostly in the acid form, and the solution is colorless. When enough NaOH solution is added to react with all of the H3O+ ions, the reaction is complete. When a small amount of extra NaOH solution is added, perhaps one drop, there will be an excess of hydroxide ions, OH−, in solution. These react with the phenolphthalein molecules, changing them from the acid form to the base form. Because the base form is red, the solution turns red, telling us that the reaction is complete (or just slightly beyond complete).
The volume of titrant added from the buret is measured. For our example, let’s assume that 18.3 mL of 0.115 M NaOH has been added. The following setup shows how the molarity of the nitric acid solution can be calculated from this data.
? mol HNO3 = 18.3 ml NaOH x 1000 ml x 0.115 mol NaOH x 1 mol HNO3
1 L HNO3 25.00 ml HNO3 1L 1000 ml NaOH 1 mol NaOH
= or 0.0842 M HNO3
The first step the unit analysis thought-process is to clearly identify the units that you want. Molarity describes the number of moles of solute per liter of solution, so we start by placing moles of HNO3 over 1 L HNO3 solution.
Because molarity is a ratio of two units, we begin our calculation with a ratio of two units. Knowing that we want volume of HNO3 solution on the bottom when we are done, we place 25.00 mL HNO3 solution on the bottom at the start. We place 18.3 mL NaOH solution on the top of our ratio, giving us the ratio of two units overall that we want.
We convert milliliters of HNO3 solution to liters of HNO3 solution using the relationship between milliliters and liters. The last two conversion factors convert from amount of one substance in a chemical reaction (mL NaOH solution) to amount of another substance in the reaction (mol HNO3). Thus this is an equation stoichiometry problem that requires at its core the conversion of moles of NaOH to moles of HNO3 using the molar ratio for the reaction between them.
NaOH(aq) + HNO3(aq) NaNO3(aq) + H2O(l)
In order to use the molar ratio to convert from moles of NaOH to moles of HNO3, we need to convert from volume of NaOH solution to moles of NaOH using the molarity as a conversion factor.
This is a basic Titration problem, the fundamentals of titration.
-Leomar T. Santiago
IV-Zara, Group5
Direct Titration
Direct titration is the simplest way among the three different ways in titration namely direct, indirect and back titration.
In direct titration. There are only 2 reactants involve and one product.
Reactants are Analyte and Titrant.
Analyte: Substance to be titrated/ solute
Titrant: Use to titrate to the analyte/ solution.
The direct titration reaction is given by:
aA + tT→ pP
Examples:
Standardization:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Analyte Titrant salt
SR: 2 mol HCl : 1 mol Na2CO3
M = g analyte
ml of solution
a.) 0.1105 x 1 mol Na2CO3 x 1000 mol = 1.04 mmol Na2CO3
105.99 1 mol
b.) 1.04 x 2 mol HCl = 2.08 mmol HCl
1 mmol Na2CO3
c.) M = 2.08 mmol HCl = 0.094 mmol/ml
22.18 ml
-Jean Ellice P. Roy
IV-Zara, Group5
In direct titration. There are only 2 reactants involve and one product.
Reactants are Analyte and Titrant.
Analyte: Substance to be titrated/ solute
Titrant: Use to titrate to the analyte/ solution.
The direct titration reaction is given by:
aA + tT→ pP
Examples:
Standardization:
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Analyte Titrant salt
SR: 2 mol HCl : 1 mol Na2CO3
M = g analyte
ml of solution
a.) 0.1105 x 1 mol Na2CO3 x 1000 mol = 1.04 mmol Na2CO3
105.99 1 mol
b.) 1.04 x 2 mol HCl = 2.08 mmol HCl
1 mmol Na2CO3
c.) M = 2.08 mmol HCl = 0.094 mmol/ml
22.18 ml
-Jean Ellice P. Roy
IV-Zara, Group5
INDIRECT TITRATION
Indirect titration requires only one reaction. Thus, this favors in product. In this reaction, reducing agent, capable of losing electrons, and the oxidizing agent, capable of gaining electrons, are kept separately and where the transfer of electrons from one species to another does not take place directly. This involves two processes: preliminary reaction and titration. Titrant does not react directly with the analyte but to a product chemically related to the analyte.
For preliminary reaction: aA + tT→ pP
For titration: pP + tT→ fF
Let’s have some examples:
Back (Indirect) Titration to Determine the Concentration of a Volatile Substance.
Example1: A child 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 25.00 mL 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
1. Write the equation for the titration:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
acid + carbonate → salt + carbon dioxide + water
2. 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
3. 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
4. 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
1. 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
2. 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
3. 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)
4. 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.
5. 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
6. The concentration of ammonia in the cloudy ammonia solution was 0.114M.
Back (Indirect) Titration to Determine the Amount of an Insoluble Salt
Example 2: 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
1. Write the equation for the titration:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
2. 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
3. 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
4. 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
1. 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
2. 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
3. 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)
4. 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.
5. 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
6. The mass of calcium carbonate in the chalk was 0.099g
-Christine Joy G. Jimenez
IV-Zara, Group5
For preliminary reaction: aA + tT→ pP
For titration: pP + tT→ fF
Let’s have some examples:
Back (Indirect) Titration to Determine the Concentration of a Volatile Substance.
Example1: A child 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 25.00 mL 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
1. Write the equation for the titration:
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
acid + carbonate → salt + carbon dioxide + water
2. 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
3. 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
4. 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
1. 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
2. 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
3. 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)
4. 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.
5. 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
6. The concentration of ammonia in the cloudy ammonia solution was 0.114M.
Back (Indirect) Titration to Determine the Amount of an Insoluble Salt
Example 2: 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
1. Write the equation for the titration:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
acid + base → salt + water
2. 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
3. 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
4. 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
1. 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
2. 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
3. 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)
4. 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.
5. 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
6. The mass of calcium carbonate in the chalk was 0.099g
-Christine Joy G. Jimenez
IV-Zara, Group5
Indirect Redox Titration
Before we start dealing with our lesson – Indirect Redox Titration, let’s go over for some scientific terms.
> 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.
> Indirect Titration - indirect titration is a process where in the analyte did not react with the titrant, directly. Instead,they are connected with the use of iodine. > Redox Titration - Redox titrations are based on a reduction-oxidation reaction between an oxidizing agent and a reducing agent.
Summing up the given definitions, we can condense that Indirect Redox Titration is a process in which the analyte did not react with the titrant directly but connected with iodine and is based on a reduction-oxidation reaction.
Example:
The Iodine produced when an excess of KI was added to a solution containing of 0.1518g K2Cr2O7 required a 46.13 mL titration with Na2S2O3. Calculate the molar concentration of the thiosulfate solution.
Cr2O7 →Cr+3
I- → I2
14H+ + 6e- + Cr2O7-2 → 2Cr+3 + 7H2O (reduction)
(2I- → I2 +2e-) 3 (oxidation)
Prelim: 14H+ +Cr2O7-2 + 6I- → 2Cr+3 + 3I2 + 7H2O
Titration: I2 + 2S2O3-2 → 2I- + S4O6-2
Preliminary; SR: 1 mole Cr2O7-2 = 6 mole I-
6 mole I- = 3 mole I2
Titration; SR: 1 mole I2 = 2 mole S2O3-2
Convert the given mass of K2Cr2O7 to the moles of K2Cr2O7 then find the mole of I-. Next, convert it to the moles of I2 then to the moles of the product - S2O3. When you find the moles of S2O3 convert it to millimoles.
(0.1518g K2Cr2O7)(1mole K2Cr2O7 / 292.4g K2Cr2O7)(6 mole I- / 1mole K2Cr2O7) (3mole I2 / 6 mole I-)(2mole S2O3 / 1mole I-)(1000mmoles S2O3 / 1 mole S2O3) = 3.096 mmol S2O3.
To find the molarity of the product - S2O3, divide its mmoles to the given amount of volume.
M S2O3 = 3.096 mmol/ 46.13mL = 0.067M S2O3
Example problem:
1. A 0.1017g sample of KbrO3 was dissolved in diluted HCl and treated with an unmeasured excess of KI. The liberated Iodine required 39.75 mL of Na2S2O3 solution. Calculate the M of Na2S2O3.
-John Bennedick B. Quijano
IV-Zara, Group5
Back Titration
Back titration - is a titration done in reverse; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess is titrated. A back titration is useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration, as with precipitation reactions. Back titrations are also useful if the reaction between the analyte and the titrant is very slow, or when the analyte is in a non-soluble solid.
Sample Problem :
A 0.4755 g sample containing ammonium oxalate (NH4)2C2O4 and inert compounds has dissolved in water and made alkaline with KOH. The liberated NH3 was distilled into 50 mL of 0.1007 N H2SO4. The excess H2SO4 was back titrated with 11.13 mL of 0.1214 N NaOH. Calculate the %N and %(NH4)2C2O4 in the sample..
Answer:
Molarity H2SO4 = 0.1007 / 2 = 0.05035 Moles H2SO4 = 50 x 0.05035 / 1000 = 0.00252 Moles NaOH = 11.13 x 0.1214 / 1000 = 0.00135 2NaOH + H2SO4 = Na2SO4 + 2H2O Moles H2SO4 titrated by NaOH = 2 x 0.00135 = 0.00270 Moles H2SO4 in excess = 0.00270 - 0.00252 = 0.00018 Moles NH3 = 2 x 0.0018 = 0.00036 the ratio between ammonium oxalate and ammonia is 2 : 1 moles ammonium oxalate = 2 x 0.00036 =
0.00072 Molar mass = 124 g/mol Grams ammonium oxalate = 0.00072 mol x 124 g/mol = 0.0893 0.0893 x 100 / 0.4755 = 18.8 % ( % ammonium oxalate in the sample) Moles N = 0.00072 x 2 = 0.00144 Grams N = 0.00144 x 14 = 0.02016 % = 0.02016 x 100 / 0.4755 = 4.24 % ( % N in the sample)
Back-titration Practice Problems
1. A 1.0000 gram sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make
250.0 mL of solution. A 25.00 mL aliquot is taken and titrated with 0.1000 M HCl:
K2CO3(aq) + 2HCl(aq) 2KCl(aq) + H2O(l) + CO2(g)
How many mL of HCl are used? (14.47)
2. A 0.6000 g sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make
200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an
Erlenmeyer flask. To the flask is added 20.00 mL of 0.1700 M HCl:
K2CO3(aq) + 2HCl(aq) 2KCl(aq) + H2O(l) + CO2(g)
The resulting solution is then titrated with 0.1048 M NaOH.
NaOH(aq) + HCl(aq) H2O(l) + NaCl(aq)
How many mL of NaOH are used? (24.16)
3. A 0.4108 gram sample of CaCO3 (100.087 g/mol) is added to a flask along with
15.00 mL of 2.000 M HCl.
CaCO3(aq) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
Enough water is then added to make 250.0 mL of solution A. A 20.00 mL aliquot of
solution A is taken and titrated with 0.1160 M NaOH.
NaOH(aq) + HCl(aq) H2O(l) + NaCl(aq)
How many mL of NaOH are used? (15.03)
-Matthew W. Garcia
IV-Zara, Group5
Sample Problem :
A 0.4755 g sample containing ammonium oxalate (NH4)2C2O4 and inert compounds has dissolved in water and made alkaline with KOH. The liberated NH3 was distilled into 50 mL of 0.1007 N H2SO4. The excess H2SO4 was back titrated with 11.13 mL of 0.1214 N NaOH. Calculate the %N and %(NH4)2C2O4 in the sample..
Answer:
Molarity H2SO4 = 0.1007 / 2 = 0.05035 Moles H2SO4 = 50 x 0.05035 / 1000 = 0.00252 Moles NaOH = 11.13 x 0.1214 / 1000 = 0.00135 2NaOH + H2SO4 = Na2SO4 + 2H2O Moles H2SO4 titrated by NaOH = 2 x 0.00135 = 0.00270 Moles H2SO4 in excess = 0.00270 - 0.00252 = 0.00018 Moles NH3 = 2 x 0.0018 = 0.00036 the ratio between ammonium oxalate and ammonia is 2 : 1 moles ammonium oxalate = 2 x 0.00036 =
0.00072 Molar mass = 124 g/mol Grams ammonium oxalate = 0.00072 mol x 124 g/mol = 0.0893 0.0893 x 100 / 0.4755 = 18.8 % ( % ammonium oxalate in the sample) Moles N = 0.00072 x 2 = 0.00144 Grams N = 0.00144 x 14 = 0.02016 % = 0.02016 x 100 / 0.4755 = 4.24 % ( % N in the sample)
Back-titration Practice Problems
1. A 1.0000 gram sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make
250.0 mL of solution. A 25.00 mL aliquot is taken and titrated with 0.1000 M HCl:
K2CO3(aq) + 2HCl(aq) 2KCl(aq) + H2O(l) + CO2(g)
How many mL of HCl are used? (14.47)
2. A 0.6000 g sample of K2CO3 (138.2055 g/mol) is dissolved in enough water to make
200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an
Erlenmeyer flask. To the flask is added 20.00 mL of 0.1700 M HCl:
K2CO3(aq) + 2HCl(aq) 2KCl(aq) + H2O(l) + CO2(g)
The resulting solution is then titrated with 0.1048 M NaOH.
NaOH(aq) + HCl(aq) H2O(l) + NaCl(aq)
How many mL of NaOH are used? (24.16)
3. A 0.4108 gram sample of CaCO3 (100.087 g/mol) is added to a flask along with
15.00 mL of 2.000 M HCl.
CaCO3(aq) + 2HCl(aq) CaCl2(aq) + H2O(l) + CO2(g)
Enough water is then added to make 250.0 mL of solution A. A 20.00 mL aliquot of
solution A is taken and titrated with 0.1160 M NaOH.
NaOH(aq) + HCl(aq) H2O(l) + NaCl(aq)
How many mL of NaOH are used? (15.03)
-Matthew W. Garcia
IV-Zara, Group5
Titration Done In Reverse
Titrations are method widely used in analytical chemistry to determine acids, bases, oxidants, reductants, metal ions, and many other species. Sometimes it is not possible to use standard titration methods. For example the reaction between determined substance and titrant can be too slow, or there can be a problem with end point determination. In such situations we often use an alternative technique to direct titration. It is called BACK TITRATION.
What is back titration? It is basically, an analytical technique in chemistry, which is performed “backwards” in the method” instead of titrating the original analyte, a volumetric solution of a reagent is added to the solution to react with the analyte, and then the excess reagent is titrated.
Addition Reaction: aA + rR → pP
Back Titration: rR + tT → pP
Mathematical Relationship
Total amount of reagent = Amount of reagent reacted with analyte + Amount of reagent back titrate with titrant
What are the properties of back titration? It reacts with the excess volume of reactant which has been left over after completing reaction with the analyte from the normal titration. The substance or solution of unknown concentration of excess intermediate reactant is made to react with known volume and concentration of intermediate reactant solution in back titration.
What are the advantages and disadvantages? First, it is useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration. Second, it is suitable when trying to work out the amount of an acid or base in a non-soluble solid. However, it needs skill and practice for effective results. It consumes time if done manually
After learning all these lessons, let's consider some examples:
* A 0.6086 g sample of primary standard grade benzoic acid, C6H5COOH, was dissolved in 60.00 ml solution of Ba(OH)2. The excess base is required 4.30 ml of 0.1203 M HCl for back titration. Calculate the concentration of Ba(OH)2 solution.
Addition reaction: 2C6H5COOH + Ba(OH)2 → Ba(C6H5COO) + 2H2O
Back titration: Ba(OH)2 + 2HCl → BaCl2 + 2H2O
Mathematical Relationship
Total amount of Ba(OH)2 = Amount of Ba(OH)2 reacted with C6H5COOH + Amount of Ba(OH)2 back titrate with HCl
Step 1
mmol of Ba(OH)2 reacted with C6H5COOH
0.6086 g C6H5COOHx(1mol/122.13g) x (1 mol/ 2 mol) x (1000 mmol/ 1 mol)
= 2.49 mmol of Ba(OH)2
Step 2
mmol of HCl = MHCl VHCl
= (0.1203 M) (4.30 ml)
= 0.52 mmol HCl
mmol of Ba(OH)2 back titrate with HCl
0.52 mmol HCl
Step 3
Total amount of Ba(OH)2 = Amount of Ba(OH)2 reacted with C6H5COOH + Amount of Ba(OH)2 back titrate with HCl
= 2.49 mmol of Ba(OH)2 + 0.26 mmol Ba(OH)2
Total amount of Ba(OH)2 = 2.75 mmol Ba(OH)2
Step 4
Calculate the concentration of Ba(OH)2 solution.
M Ba(OH)2 = (2.75 mmol Ba(OH)2/60.00 mL Ba(OH)2)
= 0.46 M Ba(OH)2
* A150.0 mL of 0.2105 M nitric acid (HNO3) was added in excess to 1.3415 g calcium carbonate (CaCO3). The excess acid was back titrated with 0.1055 M sodium hydroxide (NaOH). It required 75.5 mL of the base to reach. Calculate the percentage (w/w) of calcium carbonate in the sample.
Addition reaction: CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O
Back titration: HNO3 + NaOH → NaNO3 + H2O
Step 1.
Calculate number of millimole of HNO3.
mmol of HNO3 = M HNO3 V HNO3
= (0.2105M) (150ml)
= 31.58 mmol HNO3
Calculate number of millimole of excess NaOH.
mmol of NaOH = M NaOH V NaOH
= (0.1055M) (75.5ml)
= 7.97 mmol NaOH
mmol of HNO3 reacted with CaCO3 = total mmol of HNO3 - mmol of HNO3 back titrate with NaOH
= 31.58 mmol - 7.97 mmol
= 23.61 mmol HNO3
Step 2
23.61 mmol HNO3
Step 3
Gram CaCO3 = Mole x Molar Mass
= 11.81 x 10-3 x 100
= 1.1805 g CaCO3
Step 4
Find percentage of calcium carbonate in the sample.
% CaCO3 = (weight of CaCO3/weight of sample) x 100%
= (1.18.5/1.3415) 100 %
= 87.99 % CaCO3
Well, now that you have read the definition and gone through an example, here is one example which I think you can attempt to solve all by yourself.
A 0.6910 g sample of K2CO3 (138.20 g/mol) is dissolved in enough water to make 200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an Erlenmeyer flask. To the flask is added 20.00 mL of 0.2000 M HCl. The resulting solution is then titrated with 0.1500 M NaOH. How many mL of NaOH are used?
Addition reaction: K2CO3+ 2HCl → 2KCl + H2O + CO2
Back Titration: HCl + NaOH → H2O + NaCl
-Camille R. Villania
IV-Zara, Group 5
What is back titration? It is basically, an analytical technique in chemistry, which is performed “backwards” in the method” instead of titrating the original analyte, a volumetric solution of a reagent is added to the solution to react with the analyte, and then the excess reagent is titrated.
Addition Reaction: aA + rR → pP
Back Titration: rR + tT → pP
Mathematical Relationship
Total amount of reagent = Amount of reagent reacted with analyte + Amount of reagent back titrate with titrant
What are the properties of back titration? It reacts with the excess volume of reactant which has been left over after completing reaction with the analyte from the normal titration. The substance or solution of unknown concentration of excess intermediate reactant is made to react with known volume and concentration of intermediate reactant solution in back titration.
What are the advantages and disadvantages? First, it is useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration. Second, it is suitable when trying to work out the amount of an acid or base in a non-soluble solid. However, it needs skill and practice for effective results. It consumes time if done manually
After learning all these lessons, let's consider some examples:
* A 0.6086 g sample of primary standard grade benzoic acid, C6H5COOH, was dissolved in 60.00 ml solution of Ba(OH)2. The excess base is required 4.30 ml of 0.1203 M HCl for back titration. Calculate the concentration of Ba(OH)2 solution.
Addition reaction: 2C6H5COOH + Ba(OH)2 → Ba(C6H5COO) + 2H2O
Back titration: Ba(OH)2 + 2HCl → BaCl2 + 2H2O
Mathematical Relationship
Total amount of Ba(OH)2 = Amount of Ba(OH)2 reacted with C6H5COOH + Amount of Ba(OH)2 back titrate with HCl
Step 1
mmol of Ba(OH)2 reacted with C6H5COOH
0.6086 g C6H5COOHx(1mol/122.13g) x (1 mol/ 2 mol) x (1000 mmol/ 1 mol)
= 2.49 mmol of Ba(OH)2
Step 2
mmol of HCl = MHCl VHCl
= (0.1203 M) (4.30 ml)
= 0.52 mmol HCl
mmol of Ba(OH)2 back titrate with HCl
0.52 mmol HCl
Step 3
Total amount of Ba(OH)2 = Amount of Ba(OH)2 reacted with C6H5COOH + Amount of Ba(OH)2 back titrate with HCl
= 2.49 mmol of Ba(OH)2 + 0.26 mmol Ba(OH)2
Total amount of Ba(OH)2 = 2.75 mmol Ba(OH)2
Step 4
Calculate the concentration of Ba(OH)2 solution.
M Ba(OH)2 = (2.75 mmol Ba(OH)2/60.00 mL Ba(OH)2)
= 0.46 M Ba(OH)2
* A150.0 mL of 0.2105 M nitric acid (HNO3) was added in excess to 1.3415 g calcium carbonate (CaCO3). The excess acid was back titrated with 0.1055 M sodium hydroxide (NaOH). It required 75.5 mL of the base to reach. Calculate the percentage (w/w) of calcium carbonate in the sample.
Addition reaction: CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O
Back titration: HNO3 + NaOH → NaNO3 + H2O
Step 1.
Calculate number of millimole of HNO3.
mmol of HNO3 = M HNO3 V HNO3
= (0.2105M) (150ml)
= 31.58 mmol HNO3
Calculate number of millimole of excess NaOH.
mmol of NaOH = M NaOH V NaOH
= (0.1055M) (75.5ml)
= 7.97 mmol NaOH
mmol of HNO3 reacted with CaCO3 = total mmol of HNO3 - mmol of HNO3 back titrate with NaOH
= 31.58 mmol - 7.97 mmol
= 23.61 mmol HNO3
Step 2
23.61 mmol HNO3
Step 3
Gram CaCO3 = Mole x Molar Mass
= 11.81 x 10-3 x 100
= 1.1805 g CaCO3
Step 4
Find percentage of calcium carbonate in the sample.
% CaCO3 = (weight of CaCO3/weight of sample) x 100%
= (1.18.5/1.3415) 100 %
= 87.99 % CaCO3
Well, now that you have read the definition and gone through an example, here is one example which I think you can attempt to solve all by yourself.
A 0.6910 g sample of K2CO3 (138.20 g/mol) is dissolved in enough water to make 200.0 mL of solution A. A 20.00 mL aliquot of solution A is taken and put into an Erlenmeyer flask. To the flask is added 20.00 mL of 0.2000 M HCl. The resulting solution is then titrated with 0.1500 M NaOH. How many mL of NaOH are used?
Addition reaction: K2CO3+ 2HCl → 2KCl + H2O + CO2
Back Titration: HCl + NaOH → H2O + NaCl
-Camille R. Villania
IV-Zara, Group 5
Direct Redox Titration
Titration is the act of adding standard solution in small quantities to the test solution until the reaction is complete. Redox titration (also called oxidation-reduction titration) is a type of titration based on a redox reaction between the analyte and titrant. It is based on transfer of electron from reducing agents to oxidizing agents.
Oxidation – loss of electron
Reduction – gain of electron
Redox titration may involve the use of a redox indicator and/or a potentiometer. It has three types namely, direct, indirect and back redox titration.
In Direct titration substance is directly titrated with titrant by using simple indicator. Below is the example of a direct redox titration.
Problem : 0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium permanganate (from a buret) to titrate it and cause it to turn pink (the end point).
The equation for this reaction is:
5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10 CO2(g) + 8 H2O(l)
(a) How many moles of sodium oxalate are present in the flask?
(b) How many moles of potassium permanganate have been titrated into the flask to reach the end point?
(c) What is the molarity of the potassium permanganate?
Solution to (a):
0.2640 g / 134.00 g/mol = 0.001970149 mol
to four sig figs, this would be 0.001970 mol
Solution to (b):
1) Determine Fe(II) in solution:
(0.01625 mol/L) (0.03226 L) = 0.000524225 mol dichromate
0.000524225 mol times (6 mol Fe(II) / 1 mol dichromate) = 0.00314535 mol Fe(II)
0.00314535 mol Fe(II) times 55.845 g/mol = 0.175652 g
2) Determine percent of iron in the sample:
0.175652 g / 1.2765 g = 13.76%
Solution to (c):
Which compound contains 13.76% iron? The only way to determine this is to calculate the percent composition of the three sbstances.
Ferrous iodate:
Fe(IO3)2
% Fe: 55.845 g/mol divided by 405.67 g/mol = 13.77%
Here's the percent composition calculator I used to calculate the three percent compositions. Here are the other two substances in the question:
Fe3(PO4)2 = 46.87%
Fe(C2H3O2)2 = 32.11 %
-Crissa Chirene P. Bug-os
IV-Zara, Group 5
Oxidation – loss of electron
Reduction – gain of electron
Redox titration may involve the use of a redox indicator and/or a potentiometer. It has three types namely, direct, indirect and back redox titration.
In Direct titration substance is directly titrated with titrant by using simple indicator. Below is the example of a direct redox titration.
Problem : 0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium permanganate (from a buret) to titrate it and cause it to turn pink (the end point).
The equation for this reaction is:
5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) + 10 CO2(g) + 8 H2O(l)
(a) How many moles of sodium oxalate are present in the flask?
(b) How many moles of potassium permanganate have been titrated into the flask to reach the end point?
(c) What is the molarity of the potassium permanganate?
Solution to (a):
0.2640 g / 134.00 g/mol = 0.001970149 mol
to four sig figs, this would be 0.001970 mol
Solution to (b):
1) Determine Fe(II) in solution:
(0.01625 mol/L) (0.03226 L) = 0.000524225 mol dichromate
0.000524225 mol times (6 mol Fe(II) / 1 mol dichromate) = 0.00314535 mol Fe(II)
0.00314535 mol Fe(II) times 55.845 g/mol = 0.175652 g
2) Determine percent of iron in the sample:
0.175652 g / 1.2765 g = 13.76%
Solution to (c):
Which compound contains 13.76% iron? The only way to determine this is to calculate the percent composition of the three sbstances.
Ferrous iodate:
Fe(IO3)2
% Fe: 55.845 g/mol divided by 405.67 g/mol = 13.77%
Here's the percent composition calculator I used to calculate the three percent compositions. Here are the other two substances in the question:
Fe3(PO4)2 = 46.87%
Fe(C2H3O2)2 = 32.11 %
-Crissa Chirene P. Bug-os
IV-Zara, Group 5
STANDARDIZATION of a solution
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 asvolumetric 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 also meant as 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.
Direct titration is a type of titration that determines the contents of a substance quantitatively. Scientists may be aware of a reactant, but not know the reactant's quantity. Direct titration is sometimes based on indicators that respond to the analyzed material, called the analyte. An example of direct titration is given below.
Example:
a. Sodium carbonate is a primary standard base that reacts with Hydrochloric acid. If 40.37 ml of a solution were required to titrate a solution containing 221.4 mg of primary standard Sodium carbonate (MM = 105.99), calculate the molarity of the HCl solution.
Balanced chemical formula:
Na2CO3 + 2HCl↔ 2NaCl + CO2 + H2O
* We need a balanced chemical formula to know the relationship between the analyte and titrant so that we can formulate the data given for the analyte into titrant and solve for the molarity.
221.4mg x 1g x 1 mol Na2CO3 x 2 mol HCl x 1000 mmol
MHCl = 1000 mg 105 .99 g Na2CO3 1 mol Na2CO3 1 mol
40.37 ml HCl solution
MHCl = 0.10 M HCl solution
* In this formulation, 221.4 mg of sodium carbonate is converted into grams so that we can convert it into mol by dividing into the molar mass of sodium carbonate which is 105.99 g. Now that it’s in mol we can use the balanced chemical formula to convert it into the mol of hydrochloric acid by multiplying it in the ratio of HCl and Na2CO3 which is 2 is to 1. Dividing it 1000 millimoles will convert it into millimoles so that we can divide it to the volume given to the titrant and get the molarity of the HCl solution.
Direct titration is a type of titration that determines the contents of a substance quantitatively. Scientists may be aware of a reactant, but not know the reactant's quantity. Direct titration is sometimes based on indicators that respond to the analyzed material, called the analyte. An example of direct titration is given below.
Example:
a. Sodium carbonate is a primary standard base that reacts with Hydrochloric acid. If 40.37 ml of a solution were required to titrate a solution containing 221.4 mg of primary standard Sodium carbonate (MM = 105.99), calculate the molarity of the HCl solution.
Balanced chemical formula:
Na2CO3 + 2HCl↔ 2NaCl + CO2 + H2O
* We need a balanced chemical formula to know the relationship between the analyte and titrant so that we can formulate the data given for the analyte into titrant and solve for the molarity.
221.4mg x 1g x 1 mol Na2CO3 x 2 mol HCl x 1000 mmol
MHCl = 1000 mg 105 .99 g Na2CO3 1 mol Na2CO3 1 mol
40.37 ml HCl solution
MHCl = 0.10 M HCl solution
* In this formulation, 221.4 mg of sodium carbonate is converted into grams so that we can convert it into mol by dividing into the molar mass of sodium carbonate which is 105.99 g. Now that it’s in mol we can use the balanced chemical formula to convert it into the mol of hydrochloric acid by multiplying it in the ratio of HCl and Na2CO3 which is 2 is to 1. Dividing it 1000 millimoles will convert it into millimoles so that we can divide it to the volume given to the titrant and get the molarity of the HCl solution.
Weight/Weight Percent (w/w%): This unit of concentration is often used for concentrated solutions, typically acids and bases. If you were to look on a bottle of a concentrated acid or base solution the concentration expressed as a weigh/weight percent. A weight/weight percent is defined as:
%m/m= (mass of solute/mass of solution)x100%
Molarity (M) - one of the most common units of concentration used by the chemist, is defined as the number of moles of solute per liter of solution. It has as its unit the capital letter "M" and is read as "molar".
M=moles of solute/liters of solution
Molality (m) - another unit of concentration used by chemists, is defined as the number of moles of solute per kilogram of solvent. It has as its unit the lower case letter "m" and is read as "molal".
m= moles of solute/ mass of solvent
Dilutions- you dilute a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation:
MiVi = MfVf
where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.
Examples:
1. Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.
Solution:
20 g NaCl / 100 g solution x 100 = 20% NaCl solution
2. What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution?
Solution:
11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2
100 mL x 1 L / 1000 mL = 0.10 L
molarity = 0.10 mol / 0.10 L
molarity = 1.0 M
3. What is the molality of a solution of 10 g NaOH in 500 g water?
Solution:
10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg
molality = 0.50 m
4. How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?
Solution:
5.5 M x V1 = 1.2 M x 0.3 L
V1 = 1.2 M x 0.3 L / 5.5 M
V1 = 0.065 L
V1 = 65 mL
So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your container and add water to get 300 mL final volume.
-Lori Anne M. Pestaño
IV-Zara, Group 5
%m/m= (mass of solute/mass of solution)x100%
Molarity (M) - one of the most common units of concentration used by the chemist, is defined as the number of moles of solute per liter of solution. It has as its unit the capital letter "M" and is read as "molar".
M=moles of solute/liters of solution
Molality (m) - another unit of concentration used by chemists, is defined as the number of moles of solute per kilogram of solvent. It has as its unit the lower case letter "m" and is read as "molal".
m= moles of solute/ mass of solvent
Dilutions- you dilute a solution whenever you add solvent to a solution. Adding solvent results in a solution of lower concentration. You can calculate the concentration of a solution following a dilution by applying this equation:
MiVi = MfVf
where M is molarity, V is volume, and the subscripts i and f refer to the initial and final values.
Examples:
1. Determine the percent composition by mass of a 100 g salt solution which contains 20 g salt.
Solution:
20 g NaCl / 100 g solution x 100 = 20% NaCl solution
2. What is the molarity of a solution made when water is added to 11 g CaCl2 to make 100 mL of solution?
Solution:
11 g CaCl2 / (110 g CaCl2 / mol CaCl2) = 0.10 mol CaCl2
100 mL x 1 L / 1000 mL = 0.10 L
molarity = 0.10 mol / 0.10 L
molarity = 1.0 M
3. What is the molality of a solution of 10 g NaOH in 500 g water?
Solution:
10 g NaOH / (40 g NaOH / 1 mol NaOH) = 0.25 mol NaOH
500 g water x 1 kg / 1000 g = 0.50 kg water
molality = 0.25 mol / 0.50 kg
molality = 0.05 M / kg
molality = 0.50 m
4. How many millilieters of 5.5 M NaOH are needed to prepare 300 mL of 1.2 M NaOH?
Solution:
5.5 M x V1 = 1.2 M x 0.3 L
V1 = 1.2 M x 0.3 L / 5.5 M
V1 = 0.065 L
V1 = 65 mL
So, to prepare the 1.2 M NaOH solution, you pour 65 mL of 5.5 M NaOH into your container and add water to get 300 mL final volume.
-Lori Anne M. Pestaño
IV-Zara, Group 5
