Chapter 16, Problem 16.16QP

(a)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration of solution with $5.20$ of $pH$ has to be calculated..

Concept introduction:

• pH of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion or hydrogen ion concentration.

  $\begin{array}{l}\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{pH}\text{=}\text{-log}\left[{\text{H}}^{\text{+}}\right]\end{array}$

• Concentration of hydrogen ion $\left[{\text{H}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$
• For an acidic solution $\text{pH}\text{<7}$ and for a basic solution $\text{pH>}\text{7}$.

(b)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration of solution with $16.00$ of $pH$ has to be calculated..

Concept introduction:

• pH of a solution is the negative of the base $\text{-10}$ logarithm of the hydronium ion or hydrogen ion concentration.

  $\begin{array}{l}\text{pH}\text{=}\text{-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{pH}\text{=}\text{-log}\left[{\text{H}}^{\text{+}}\right]\end{array}$

• Concentration of hydrogen ion $\left[{\text{H}}^{\text{+}}\right]\text{=}{\text{10}}^{\text{-pH}}$
• For an acidic solution $\text{pH}\text{<7}$ and for a basic solution $\text{pH>}\text{7}$.

(c)

Interpretation Introduction

Interpretation:

The hydrogen ion concentration of solution with $\text{3}\text{.7×}{\text{10}}^{\text{-9}}\text{M}$ of hydroxide concentration has to be calculated

Concept introduction:

If equilibrium is established between hydronium ions, hydroxide ions and water molecules, an equilibrium law can be written as follows,

  $\begin{array}{c}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}\\ K_a\times \left[H_{2}O\right]\left[H_{2}O\right]=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ K_w=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\end{array}$

Ionic product water can also be written as,

$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]=1.0\times {10}^{-14}$