Chapter 16, Problem 12PS

### 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

# What is the pH of a 1.2 × 10˗4 M solution of KOH? What is the hydronium ion concentration of the solution?

Interpretation Introduction

Interpretation:

For a given concentration of Potassium hydroxide (KOH) solution,pH and the concentration of hydronium ion in the solution has to be calculated.

Concept introduction:

A strong base completely dissociates into its constituent ions in aqueous solution, as a result, the concentration of its ion is same as the initial concentration of that strong base.

For example, strong base dissociates as follows in water,

B(aq)+H2O(l)BH+(aq)+OH(aq)

[BH+]=[OH]=[B]

The pOH of a solution is basically the measure of the molar concentration of the OH ion in the solution. More the concentration of OH ion in the solution lower will be the value of pOH and more basic will be the solution.

The expression for pOH is given as,

pOH=log[OH]

pH is calculated using following relation.

pH+pOH=14

The concentration of H3O+ ions is calculated by using the ionic product of water.

Kw=[H3O+][OH]

The value of Kw is 1.0×1014.

If pH<7 then, the solution is acidic in nature.

If pH >7 then, the solution is basic in nature.

If pH=7 then, the solution is neutral in nature.

Explanation

The value of pH for Potassium hydroxide (KOH) solution and the concentration of hydronium ion in the solution is calculated below.

Given:

The initial concentration of KOH solution is 1.2Ã—10âˆ’4Â M.

KOH is a strong base and it dissociates as follows in water,

â€‚Â KOH(aq)+H2O(l)â‡ŒK+(aq)+OHâˆ’(aq)

The ICE table is as follows,

EquilibriumKOH(aq)+H2O(l)â‡ŒK+(aq)+OHâˆ’(aq)Initial(M)1.2Ã—10âˆ’4Â 00Change(M)+1.2Ã—10âˆ’4Â +1.2Ã—10âˆ’4Â Equilibrium(M)1.2Ã—10âˆ’4Â 1.2Ã—10âˆ’4Â

From the table,

[OHâˆ’]=1.2Ã—10âˆ’4Â Â M

The expression for pOH is given as,

â€‚Â pOH=âˆ’log[OHâˆ’]

Substitute 1.2Ã—10âˆ’4Â M for [OHâˆ’]

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