   Chapter 14, Problem 173CP

Chapter
Section
Textbook Problem

# Consider 50.0 mL of a solution of weak acid HA (Ka = 1.00 × 10−6), which has a pH of 4.000. What volume of water must be added to make the pH = 5.000?

Interpretation Introduction

Interpretation: The volume of a weak acid HA , its acid dissociation constant and pH value is given. The volume of water added in HA to change the pH from 4.000 to 5.000 is to be calculated.

Concept introduction: The measure of concentration of hydrogen ions is known as pH .

The process of reduction of concentration of a solute in a solution is known as dilution.

Explanation

Explanation

To determine: The volume of water added in HA to change the pH from 4.000 to 5.000 .

The concentration of [H3O+]  at pH=4.000 is 1.0×104M_ .

Given

Initial volume of HA is 50.0mL .

The initial pH value of HA at 50.0mL is 4.000 .

The final pH value of HA after addition of water is 5.000 .

Formula

The pH value is calculated using the formula,

pH=log10[H3O+] (1)

Where,

• [H3O+] is concentration of hydronium ions.

Substitute the initial pH value of HA in the above equation.

pH=log10[H3O+]4.000=log10[H3O+][H3O+]=1.0×104M_

The concentration of HA at pH=4.000 is 0.0101M_ .

The concentration of [H3O+] is 1.0×104M .

The value of Ka is 1.0×106 .

It is assumed that the initial concentration of HA is x .

Make the ICE table for the dissociation reaction of HA .

HA(aq)+H2O(l)H3O+(aq)+A(aq)Initial:x00Change:104104104Equilibrium:x104104104

The equilibrium constant for the given reaction is,

Ka=[H3O+][A][HA]

Substitute the equilibrium concentration values in the above expression.

Ka=[H3O+][A][HA]1.0×106=(104)2(x104)x=0.0101

Thus, initial concentration of HA at pH=4.000 is 0.0101M_ .

The concentration of [H3O+] at pH=5.000 is 1.0×105M .

Given

Initial volume of HA is 50.0mL .

The initial pH value of HA at 50.0mL is 4.000 .

The final pH value of HA after addition of water is 5.000 .

Substitute the pH value of HA in equation (1).

pH=log10[H3O+]5

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