Chapter 14, Problem 104CWP

(a)

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution after the addition of $20.00\text{mL}$ of $0.250\text{M}$ $\text{NaOH}$ is to be calculated. To make the $\text{pH}\text{7}\text{.00}$ volume of the $\text{NaOH}$ is to be calculated..

Concept introduction: $\text{HI}$ is a strong acid and NaOH is a strong base. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as $\text{pH}$ of the solution.

It is the negative logarithm of Hydrogen ion concentration.

The number of moles calculated by taking the volume in milliliter is known as number of milli moles.

$\text{pH-7}$ is the neutralization condition of every solution.

To determine: The $\text{pH}$ of the solution after the addition of $20.00\text{mL}$ of $\text{NaOH}$.

(b)

Interpretation Introduction

Interpretation: The $\text{pH}$ of the solution after the addition of $20.00\text{mL}$ of $0.250\text{M}$ $\text{NaOH}$ is to be calculated. To make the $\text{pH}\text{7}\text{.00}$ volume of the $\text{NaOH}$ is to be calculated..

Concept introduction: $\text{HI}$ is a strong acid and NaOH is a strong base. The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as $\text{pH}$ of the solution.

It is the negative logarithm of Hydrogen ion concentration.

The number of moles calculated by taking the volume in milliliter is known as number of milli moles.

$\text{pH-7}$ is the neutralization condition of every solution.

To determine: The volume of the $\text{NaOH}$ added to make the $\text{pH-7}$ in the given solution.