   Chapter 16, Problem 59PS

Chapter
Section
Textbook Problem

Calculate the pH of a 0.0010 M aqueous solution of HF.

Interpretation Introduction

Interpretation:

The pH of a 0.0010 M aqueous solution of HF has to be calculated.

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 acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Explanation

HF reacts with water. The equilibrium chemical reaction is as follows.

HF(aq) + H2O(l) F-(aq) + H3O+(aq)

The equilibrium expression:

ka[F-][H3O+][HF]

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

HF(aq) + H2O(l) F-(aq) + H3O+(aq)   I         0.12                --           --                   --C          -x                  --         +x                  +xE     (0.0010 - x)       --          x                     x

Ka of HF  is 7.2 × 10-4

At equilibrium, x = [H3O+] = [F-]

ka[F-][H3O+][HF]

7.2×10-4 = (x)(x)0.0010- x7.2×10-4 = (x)20.0010- x0

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