Chapter 17, Problem 22PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Determine the volume (in mL) of 1.00 M HC1 that must be added to 750 mL of 0.50 M HPO42− to produce a buffer with a pH of 7.00.

Interpretation Introduction

Interpretation:

Volume of 1.0 M HCl must be added to 750 mL of 0.05 M HPO42 to get the buffer solution of pH equals to 7 has to be calculated.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, HCl is titrated against NaOH.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, HCl is titrated against NH4OH.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, CH3COOH is titrated against NaOH.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, CH3COOH is titrated against NH4OH.

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 CH3COOH with NaOH. The equilibrium can be represented as,

CH3COOH(aq)+NaOH(aq)H2O(l)+CH3COONa(aq)

Calculation of pH at various points is done as follows,

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

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

(2) The pH calculation just before the equivalence point,

As the addition of NaOH is done there will be formation of buffer solution CH3COOH/CH3COO. The pH calculation for buffer solution is done by using Henderson-Hesselbalch equation.

pH=pKa+log[conjugatebase][acid] (2)

At the midpoint of the titration, when concentration of acid and its conjugate base are equal pH value at midpoint will be given as;

pH=pKa+log[conjugatebase][acid]

Substitute [conjugatebase]for[acid].

pH=pKa+log[conjugatebase][conjugatebase]=pKa+log(1)=pKa+0=pKa

Therefore, pH value at midpoint is equal to pKa.

(3) The pH calculation the equivalence point.

At equivalence point all the acid will be neutralized, and there will be only OH ion and CH3COO. The OH will be produced due to the hydrolysis of acetate ion at equivalence point. The hydrolysis equilibrium is represented as,

CH3COO(aq)+H2O(l)OH(aq)+CH3COOH(aq)

By using the value of Kb for the acetate ion, concentration of  OH can be calculated.  Thus the value of pH is greater than 7 at equivalence point for the weak acid- strong base titrations.

The relation between Ka and Kb for weak acid and its conjugate base is given as,

Kw=(Ka)(Kb) (3)

(4) The pH calculation after the equivalence point.

After the equivalence point there will be excess of OH ion in the solution and there will be very less amount of CH3COO ion. The amount of CH3COO produce can be neglected with respect to excess amount of OH.

Concentration of OH after equivalence point will be calculated by using the expression,

concentration=numberofmolestotalvolumeofsolution (4)

Explanation

The reaction between HCl and HPO42âˆ’ is actually the titration between strong acid and weak base. Therefore, the given value of pH is calculated at a point between after starting point and before equivalence point. The volume of volume of HCl used is calculated below.

Given:

Refer to table 16.2 in the textbook for the value of Kb.

The value of Kb for HPO42âˆ’ is 1.6Ã—10âˆ’7.

The pKb value is calculated as follows;

pKb=âˆ’log(Kb)

Substitute, 1.6Ã—10âˆ’7 for Kb.

pKb=âˆ’log(1.6Ã—10âˆ’7)=6.80

Therefore, pKb value is 6.80.

The value of pH is for the given solution is 7.0.

The pOH value for the given solution is calculated by busing expression, pHÂ +Â pOHÂ =Â 14

Rearrange for pOH,

Â pOHÂ =Â 14âˆ’pHÂ

Substitute 7.0 for pH

Â pOHÂ =Â 14âˆ’7=7

Therefore, the value of pOH for the given solution is 7.

The initial concentration of HPO42âˆ’ is 0.50â€‰molâ‹…Lâˆ’1.

The initial concentration of HCl is 1.00â€‰molâ‹…Lâˆ’1.

The volume of HPO42âˆ’ is 750â€‰mL.

Conversion of 750â€‰mL into L.

(750â€‰mL)(1â€‰L1000â€‰mL)=0.750â€‰L

Let the volume of HCl added xâ€‰mL.

Conversion of xâ€‰mL into L.

(xâ€‰mL)(1â€‰L1000â€‰mL)=0.00xâ€‰L

The total volume after the reaction is calculated as,

totalâ€‰volume=volumeâ€‰ofâ€‰HPO42âˆ’(L)Â +Â volumeÂ ofÂ HCl(L)

totalâ€‰volumeÂ =Â 0.750(L)+0.00x(L)=(0.750+0.00x)L

Therefore, total volume after reaction is (0.750+0.00x)â€‰L.

The calculation of moles is done by using the expression,

Numberâ€‰ofÂ moles=concentration(molâ‹…Lâˆ’1)â‹…volume(L)

In aqueous solution HCl undergoes complete ionization and the reaction is given as;

HClÂ (aq)+H2O(l)â†’H3O+(aq)+Clâˆ’(aq)

The H3O+ ions from ionization of HCl react with the weak base HPO42âˆ’. Â The concnetration of H3O+ will be equal to the concentration of HCl.

The ICE table (1) for the reaction between HCl and HPO42âˆ’ is given below,

EquationHPO42âˆ’(aq)+H3O+(aq)â†’H2O(l)+H2PO4âˆ’(aq)Initial(mol)0.375(0.00x)0Change(mol)âˆ’0.00xâˆ’0.00x0.00xAfterâ€‰reaction(mol)(0.375âˆ’0.00x)00

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started