   Chapter 14, Problem 93E

Chapter
Section
Textbook Problem

# What mass of KOH is necessary to prepare 800.0 mL of a solution having a pH = 11.56?

Interpretation Introduction

Interpretation: The mass of KOH that is necessary to prepare 800.0mL of a solution that has a pH value of 11.56 is to be calculated.

Concept introduction: The pH of a solution is define as a figure that expresses the acidity of the alkalinity of a given solution. A logarithmic scale is used on which, the value 7 corresponds to a neutral species, a value less than 7 corresponds to an acid and a value greater than 7 corresponds to a base.

The pOH of a solution is calculated by the formula, pOH=log[OH]

The sum, pH+pOH=14

Explanation

Explanation

To determine: The mass of KOH that is necessary to prepare 800.0mL of a solution that has a pH value of 11.56 .

The pOH value is 2.44_ .

Given

pH=11.56

The sum, pH+pOH=14

Substitute the given value of pH in the above expression.

11.56+pOH=14pOH=2.44_

The [OH] is 0.0036M_ .

The pOH of a solution is calculated by the formula,

pOH=log[OH]

Rearrange the above expression to calculate the value of [OH] .

[OH]=10pOH

Substitute the given value of pOH in the above expression.

[OH]=102.44=0.0036M_

The concentration of KOH is 0.0036M_ .

KOH is a strong base. It dissociates completely into its ions. It is a monoacidic base.

Therefore, the concentration of KOH is equal to the [OH] , that is 0.0036M_ .

The number of moles of KOH required is 2.928×10-3mol_

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