   Chapter 15, Problem 97AE

Chapter
Section
Textbook Problem

# A student intends to titrate a solution of a weak monoprotic acid with a sodium hydroxide solution but reverses the two solutions and places the weak acid solution in the buret. After 23.75 mL of the weak acid solution has been added to 50.0 mL of the 0.100 M NaOH solution, the pH of the resulting solution is 10.50. Calculate the original concentration of the solution of weak acid.

Interpretation Introduction

Interpretation:

The titration of a weak monoprotic acid with a sodium hydroxide solution, with the weak acid present in the buret, is given. The pH value, after a stated amount of the weak acid has been added to 50.0mL of the NaOH solution, is 10.50 . The original concentration of the solution of the weak acid is to be calculated.

Concept introduction:

A technique in which the concentration of an unknown solution is calculated by using a solution of a known concentration is known as titration.

Explanation

Explanation

To find the [OH]

The given value of pH is 10.50 .

The value of pOH is calculated by the formula,

14pH=pOH

Substitute the given value of pH in the above expression.

pOH=1410.5=3.5

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 value of pH in the above expression.

[OH]=103.5=3.16×10-4M_

To find the moles of H+

The [OH] is 3.16×104M .

The total volume =(50+23.75)mL=73.75mL=0.07375L

The value of [OH] is also calculated by the formula,

[OH]=MolesofOHTotalvolume(L)

Substitute the value of [OH] and the total volume in the above expression.

3.16×104=MolesofOH0.07375LMolesofOH=2.33×105mol

Therefore, the moles of OH remaining is 2

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