Chapter 16.5, Problem 16.10P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ for $5\times {10}^{-5}M{\text{ HNO}}_3$ has to be calculated.

Concept Introduction:

The ionization of water is as follows:

  ${\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}^{+}+{\text{OH}}^{-}$

The equilibrium constant for the reaction is called ionic product constant for water $K_{\text{w}}$. The expression for $K_{\text{w}}$ is as follows:

  $K_{\text{w}}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$

Thus, value of $K_{\text{w}}$ is ${10}^{-14}$ for pure water at $25°\text{C}$.

$\text{pH}$ is defined as negative logarithm of concentration of ${\text{H}}^{+}$. It is calculated as follows:

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

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $2\times {10}^{-6}M\text{ KOH}$ has to be calculated.

Concept Introduction:

Refer to part (a).