Chapter 15, Problem 15.33QE

(a)

Interpretation Introduction

Interpretation:

The hydroxide ion concentration when $\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]$ is $4.5\times {10}^{-4}M$ has to be determined.

Concept Introduction:

Water is a compound that can act as an acid or a base. A water molecule can donate a proton as well as it accepts a proton. Water molecules can react with other molecule of water to produce hydronium ion and hydroxide ion. This reaction is called as autoionization of water. In this reaction, water can be an acid or base.

The reaction of autoionization of water is as follows:

  ${\text{H}}_{\text{2}}\text{O}+{\text{H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{OH}}^{-}$

The equilibrium constant for autoionization of water is denoted by $K\text{'}$.

The expression for $K\text{'}$ is as follows:

    $K\text{'}=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]}{{\left[{\text{H}}_{\text{2}}\text{O}\right]}^2}$        (1)

Also, water is pure liquid. Its concentration does not appear. Therefore, the equilibrium constant $K\text{'}$ will become $K_{\text{w}}$ for pure water.

The expression for $K_{\text{w}}$ is as follows:

    $K_{\text{w}}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (2)

Also at $25°\text{C}$, $K_{\text{w}}$ is equal to $1.0\times {10}^{-14}$. The equation (2) will be modified as follows:

  $1.0\times {10}^{-14}=\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{OH}}^{-}\right]$        (3)

(b)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration when $\left[{\text{OH}}^{-}\right]$ is $8.33\times {10}^{-5}M$ has to be determined.

Concept Introduction:

Refer to part (a)