   # The equilibrium constant for the reaction of formic acid and sodium hydroxide is 1.8 × 10 10 (page 726). Confirm this value. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 103GQ
Textbook Problem
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## The equilibrium constant for the reaction of formic acid and sodium hydroxide is 1.8 × 1010 (page 726). Confirm this value.

Interpretation Introduction

Interpretation:

The equilibrium constant of the reaction is 1.8×1010 has to be proved.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

### Explanation of Solution

A strong base like NaOH, dissociates into Na+ and OH- ions. These hydroxide ions are react with HCO2H.

The equilibrium reaction is as follows.

HCO2H(aq) + OH-(aq) H2O(l) + HCO2-(aq)

Equilibrium expression:

K=[HCO2-][HCO2H][OH-]=1.8×1010                             (1)

H2O(l)H3O+(aq) + OH-(aq)

Equilibrium expression:

Kw=[H3O+][OH-]=1×10-14                                  (2)

HCO2H(aq) + H2O(l)HCO2-(aq) + H3O+(aq)

Equilibrium expression:

Ka=[HCO2-][H3O+][HCO2H]=1

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