   Chapter 17, Problem 28PS

Chapter
Section
Textbook Problem

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 1.00 × 102 mL of solution and then titrate the solution with 0.108 M NaOH.C6H5CO2H(aq) + OH−(aq) ⇄ C6H5CO2−(aq) + H2O(ℓ) (a) What was the pH of the original benzoic add solution? (b) What are the concentrations of all of the following ions at the equivalence point: Na+, H3O+, OH−, and C6H5CO2−? (c) What is the pH of the solution at the equivalence point?

a)

Interpretation Introduction

Interpretation:

The value of pH has to be calculated at the various points during the titration between C6H5COOH and NaOH. The value of pH, of the original solution of C6H5COOH.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C6H5COOH with NaOH is represented as,

C6H5COOH(aq)+ OH(aq)H2O(l)+C6H5COONa(aq)

Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

Explanation

The pH value before the titration can be calculated by using Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq)

Given:

Refer to the Apendix H in the textbook for the value of Ka.

The value of Ka for benzoic acid is 6.3×105.

The value of Kw for water is 1.0×1014.

The amount of C6H5COOH dissolved is 0.235 g.

Molar mass of C6H5COOH l is 122 gmol1.

Number of moles of C6H5COOH are calculated by using following expression,

Number of moles = weight of solute molar mass of solute(mol)

Substitute 0.235 g for weight of solute , 122 gmol1 for molar mass of solute.

Number of moles = 0.235 g122 gmol1=0.0019 mol

Therefore Number of moles of C6H5COOH are 0.0019 mol.

The  volume of the solvent is 0.100 L.

The initial concentration of C6H5COOH is calculated by using the expression,

Molarity=Number of molesvolume of the solvent(molL1)

Substitute 0.0019 mol for Number of moles, 0.100 L for volume of the solvent.Molarity=0.0019 mol0.100 L=0.019 molL1

Therefore initial concentration of C6H5COOH is 0.019 molL1.

ICE table (1) gives the dissociation of C6H5COOH.

EquationC6H5COOH(aq)+H2O(l)H3O+(aq)+C6H5COO(aq)Initial(molL1)0.01900Change(molL1)x+x+xAfterreaction(molL1)0

(b)

Interpretation Introduction

Interpretation:

The value of pH is to be calculated at the various points during the titration between C6H5COOH and NaOH. The concentration of OH,H3O+,C6H5COO and Na+ has to be calculated at equivalence point.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C6H5COOH with NaOH is represented as,

C6H5COOH(aq)+ OH(aq)H2O(l)+C6H5COONa(aq)

Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

(c)

Interpretation Introduction

Interpretation:

The value of pH has to be calculated at the various points during the titration between C6H5COOH and NaOH. The value of pH at equivalence point has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution.

For weak acid-strong base titration the pH value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of C6H5COOH with NaOH is represented as,

C6H5COOH(aq)+ OH(aq)H2O(l)+C6H5COONa(aq)

Ka and its relation with H3O+ ion concentration.

Ka=[H3O+](eq)[A](eq)[HA](eq) (1)

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