   Chapter 17.2, Problem 17.2CYU

Chapter
Section
Textbook Problem

What is the pH of a buffer solution composed of 0.50 M formic acid (HCO2H) and 0.70 M sodium formate (NaHCO2)?

Interpretation Introduction

Interpretation:

The value of pH has to be calculated for the formic acid/ sodium formate buffer solution.

Concept introduction:

A buffer solution is defined as a solution which does not show any change in its pH value on the addition of small amount of acid or base.There are two types of buffer solutions, (1) Acidic Buffer and (2) Basic Buffer.

Acidic Buffer is a solution which has the pH value below 7. An acidic buffer is prepared by mixing a weak acid with its conjugate base. For example a mixture of weak acid CH3COOH and its conjugate base CH3COO gives a acidic buffer.

CH3COOH(aq.)+ H2O(l)H3O+(aq.)+ CH3COO1(aq.)

Basic Buffer is a solution which has the pH value above 7. A basic buffer is prepared by mixing a weak base with its conjugate acid. For example a mixture of weak base NH4OH and its conjugate acid NH4+ gives a baisc buffer.

NH4OH(aq.)NH4+(aq.)+ OH1(aq.)

Explanation

The pH value calculation for the given solution of foemic acid/sodium formate is given below.

Given:

Refer to table 16.2 in the text book for the value of Ka.

The value Ka for formic acid is 1.8×104.

The value of pKa is calculated as follows;

pKa=log(Ka)

Substitute, 1.8×104 for Ka.

pKa=log(1.8×104)=3.74

Therefore, pKa value of formic acid is 3.74.

The value of concentration for formic acid solution is 0.50molL1.

The value of concentration for sodium formate solution is 0.70molL1.

There will be equilibrium between formic acid and sodium formate which can be represented in ICE table as follows,

EquationHCOOH(aq)+H2O(aq)H3O+(aq)+HCOO(aq)Initial(M)0.5000.70Change(M)x+x+xEquilibrium(M)0.50xx0.70+x

The acid dissociation constant expression is given as;

Ka=[H3O+](eq)[HCOO](eq)[HCOOH](eq) (1)

Rearrange the equation (1) for [H3O+] .

[H3O+](eq)=[HCOOH](eq)(Ka)[HCOO](eq)                                                                   (3)

From the ICE table,

[H3O+](eq)=xmolL1[HCOO](eq)=(0

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