Chapter 17, Problem 8PS

### 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

# Lactic acid (CH3CHOHCO2H) is found in sour milk, in sauerkraut, and in muscles after activity. (Ka for lactic acid = 1.4 × 10−4.) (a) If 2.75 g of NaCH3CHOHCO2, sodium lactate, is added to 5.00 × 102 mL of 0.100 M lactic acid, what is the pH of the resulting buffer solution? (b) Is the pH of the buffered solution lower or higher than the pH of the lactic acid solution?

a)

Interpretation Introduction

Interpretation:

Calculate the pH of the resulting buffer solution formed by adding 2.75 g of NaCH3CHOHCO2, sodium lactate, to 5.00×102 ml of 0.100 M lactic acid.

Concept introduction:

The Henderson-Hasselbalch equation relates pH of a buffer to pKa of acid, concentration of conjugate base and concentration of acid. The expression is written as,

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

This equation shows that pH of the buffer solution is controlled by two major factors. First, Strength of the acid can be expressed in terms of pKa and second, the relative concentration of acid and its conjugate base at equilibrium. It can be seen from equation (1) that pH of the buffer solution is comparable to pKa values. So this equation can be used to establish a relation between pH and pKa value of acid.

pKa is defined as the negative logarithm of Ka value of the acid and is given as

pKa=log(Ka) (2)

Explanation

The calculation of pH is done by using Henderson-Hasselbalch equation.

Given:

Acid dissociation constant, Ka for lactic acid is 1.4Ã—10âˆ’4.

The concentration of lactic acid is 0.100Â M.

The volume of the solution is 5.00Ã—102Â ml.

Amount of sodium lactate added is 2.75â€‰g.

Lactic acid undergoes dissociation as follows,

â€‚Â CH3CHOHCO2H(aq)â‡ŒCH3CHOHCO2âˆ’(aq)+H+(aq)

Sodium lactate ion, NaCH3CHOHCO2, is the conjugate base of the lactic acid, CH3CHOHCO2H.

The concentration of sodium lactate is calculated as follows,

[NaCH3CHOHCO2]=wMVÂ  (3)

Here,

• w is the given mass of sodium lactate.
• M is the gram molecular mass of the sodium lactate.
• VÂ  is the volume of the solution in liters.

The volume of the solution is,

VÂ =(5.00Ã—102Â ml)(1.00Â L1000Â ml)=0.5Â L

Gram molecular mass of sodium lactate is 112.06Â gâ‹…mol.

Substitute 2.75â€‰g for w, 112

b)

Interpretation Introduction

Interpretation:

Calculate the value of pH for 0.1 M solution of lactic acid and then compare with the pH of the buffered solution.

Concept introduction:

In aqueous solution an acid undergoes ionization. The ionization of an acid is expressed in terms of the equilibrium constant. The quantitative measurement tells about the strength of the acid. Higher the value of Ka stronger will be the acid. The acid dissociation can be represented as following equilibrium

HA(aq.)+ H2O(l)H3O+(aq.)+ A1(aq.)

A weak acid undergoes partial dissociation in aqueous solution and the expression for dissociation constant,Ka (or equilibrium constant) is given as,

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

Here,

• [H3O+](eq) is the equilibrium concentration of hydronium ion.
• [A](eq) is the equilibrium concentration of conjugate base of the acid.
• [HA](eq) is the equilibrium concentration of acid.

The ICE table gives the relationship between the concentrations of species at equilibrium.

EquationHA(aq)+H2O(aq)H3O+(aq)+AInitial(M)c00Change(M)cx+cx+cxEquilibrium(M)ccxcxcx

The expression for the acid-dissociation constant, Ka, is given as follows,

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

The pH of a solution is basically the measure of the molar concentration of the H+ or H3O+ ion in the solution. More the concentration of H+ or H3O+ ion in the solution, lesser will be the pH value and more acidic will be the solution.

The expression for pH is given as,

pH=log[H+] (4)

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