Chapter 15, Problem 71AE

Interpretation Introduction

Interpretation:

Volume of $0.60M\text{ HCl}$ is required to be added to neutral sodium rhodizonate to form $500\text{ mL}$ of solution with $\text{pH}$ of 2.8 has to be calculated.

Concept Introduction:

The acidity strength is quantified by the magnitude of a parameter called $\text{pH}$. The strong acids tend to lower values of $\text{pH}$. This is since $\text{pH}$ represents the negative logarithm of the concentration of ${\text{H}}^{+}$ ion. So higher the concentration of ${\text{H}}^{+}$ ion lower will be $\text{pH}$.

The expression to evaluate $\text{pH}$ is as follows:

  $\text{pH}=-\log \left[{\text{H}}^{+}\right]$

Here,

$\left[{\text{H}}^{+}\right]$ denotes concentration of ${\text{H}}^{+}$ ions.

The dilution formula is given as follows:

  $M_1{V}_1=M_2{V}_2$

Here,

$M_1$ denotes molarity of  initial solution before dilution.

$V_1$ denotes the volume before dilution.

$M_2$ denotes molarity of final solution after dilution.

$V_2$ denotes volume after dilution.