Chapter 14, Problem 80QAP

Interpretation Introduction

(a)

Interpretation:

The molarity of a solution of HClO₄ of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.

Concept introduction:

The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.

Interpretation Introduction

(b)

Interpretation:

The molarity of strong acid is to be determined.

Concept introduction:

The molarity can be calculated as follows:

$\text{molarity=}\frac{\text{number of moles of solute }}{\text{volume of solution(L)}}$

The relation between pH and pOH is as follows:

$pH+pOH=14$

The pOH can be calculated as follows:

$pOH=-{\log }_{10}[O{H}^{-}]$

Interpretation Introduction

(c)

Interpretation:

The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.

Concept introduction:

The formula which will be used is −

$M_1{V}_1=M_2{V}_2$

Here, M is molarity and V is volume.