Chapter 17, Problem 17.6QP

Determine the pH of (a) a 0.20 M NH₃ solution, and (b) a solution that is 0.20 M NH₃ and 0.30 M NH₄Cl.

Interpretation Introduction

Interpretation:

By using the given concentration of acetic acid and sodium acetate, calculate the $\text{pH}$ values.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the hydrogen ion concentration in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.

  ${\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

To calculate: the $\text{pH}$ of ammonia

Answer to Problem 17.6QP

$\text{The pH of ammonia solution is }11.28$

Explanation of Solution

The given concentrations are recorded as shown.

$\text{The}{\text{concentration of NH}}_{\text{3}}{\text{ and NH}}_{\text{4}}\text{Cl}\text{are}\text{0}\text{.20M}\text{and}\text{0}\text{.30M}$

Here acetic acid is a weak base, the equilibrium table is as follows

$\begin{array}{l}{\text{NH}}_{\text{3}}\text{(aq) +}{\text{H}}_{\text{2}}\text{O(l)}\rightleftarrows {\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq) +}{\text{OH}}^{\text{-}}\text{ (aq)}\\ \\ \text{Initial concentration(M): }\text{0}\text{.20 }\text{0}\text{.00 }\text{0}\text{.00}\\ \text{Change in concentration (M): }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{(0}\text{.20-x) }\text{x }\text{x}\end{array}$

${\text{K}}_{\text{b}}{\text{ value for NH}}_{\text{3}}\text{ is 1}\text{.8 ×}{\text{10}}^{\text{-5}}$

$\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}\\ \text{1}\text{.8}{\text{×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.20-x)}}\approx \frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \text{the}\text{value}\text{of}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.9 ×}{\text{10}}^{\text{-3}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(1}\text{.9 ×}{\text{10}}^{\text{-3}}\text{)}\\ \text{=}\text{2}\text{.72}\\ \text{pH}\text{=}\text{14}\text{-2}\text{.72}\\ \text{=}\text{11}\text{.28}\end{array}$

The ratio of concentrations of ammonium ion and hydroxide ion to concentration of ammonia is ${\text{K}}_{\text{b}}$.  By substituting the concentrations from equilibrium table, we can find the hydroxide ion concentration.  The pOH value can be calculated by taking negative logarithm of concentration of hydroxide ion.

Interpretation Introduction

Interpretation:

By using the given concentration of acetic acid and sodium acetate, calculate the $\text{pH}$ values.

Concept introduction:

• $\text{pH}$ is the logarithm of the reciprocal of the hydrogen ion concentration in a solution.
•   $\text{pH}$ is used to determine the acidity or basicity of an aqueous solution.

  ${\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

To calculate: the $\text{pH}$ of solution contains ammonia and ammonium chloride.

Answer to Problem 17.6QP

  $\text{The pH of a solution contains ammonia and ammonium chloride is }9.08$

Explanation of Solution

$\begin{array}{l}{\text{NH}}_{\text{4}}\text{Cl(aq) }\to {\text{NH}}_{\text{4}}{}^{\text{+}}{\text{(aq) + Cl}}^{\text{-}}\text{ (aq)}\\ {\text{NH}}_{\text{3}}\text{(aq) +}{\text{H}}_{\text{2}}\text{O(l)}\rightleftarrows {\text{NH}}_{\text{4}}{}^{\text{+}}\text{(aq) +}{\text{OH}}^{\text{-}}\text{ (aq)}\\ \\ \text{Initial concentration(M): }\text{0}\text{.20 }\text{0}\text{.30 }\text{0}\\ \text{Change in concentration (M): }\text{-x }\text{+x }\text{+x}\\ \text{Equilibrium}\text{concentration (M): }\text{(0}\text{.20-x) }\text{0}\text{.30}\text{+x }\text{+x}\\ \text{the concentration of ammonium should be added, because ammonium chloride fully ionizes }\\ \text{and forms equlibrium with ammonia}\end{array}$

$\begin{array}{l}{\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}\\ \text{1}\text{.8}{\text{×10}}^{\text{-5}}\text{=}\frac{\text{(x)(0}\text{.30 + x)}}{\text{(0}\text{.20 - x)}}\approx \frac{\text{x(0}\text{.30)}}{\text{0}\text{.20}}\\ \text{the}\text{value}\text{of}\text{x}\text{is}\text{very}\text{small}\text{and}\text{neglect}\text{it,}\\ \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{]}\text{=}\text{1}\text{.2 ×}{\text{10}}^{\text{-5}}\text{M}\\ \text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}\\ \text{=}\text{-log(1}\text{.2 ×}{\text{10}}^{\text{-5}}\text{M)}\\ \text{=}\text{4}\text{.92}\\ \text{pH}\text{=}\text{14}\text{-}\text{4}\text{.92}\\ \text{=}9.08\end{array}$

Another method to calculate $\text{pH}$, by using Henderson-Hasselbalch equation

$\begin{array}{l}\text{pH}\text{=}{\text{pK}}_{\text{a}}\text{+}\text{log}\frac{{\text{[NH}}_{\text{3}}\text{]}}{{\text{[NH}}_{\text{4}}{}^{\text{+}}\text{]}}\\ \text{pH}\text{=}\text{-log(5}\text{.6}\text{×}{\text{10}}^{\text{-10}}\text{)}\text{+}\text{log}\frac{\text{0}\text{.20}\text{M}}{\text{0}\text{.30}\text{M}}\\ \text{=}\text{-log(5}\text{.6}\text{×}{\text{10}}^{\text{-10}}\text{)}\text{+}\text{log}\text{0}\text{.66}\\ \text{=}9.25\text{-}\text{0}\text{.18}\\ \text{=}9.07\end{array}$

The ratio of concentrations of ammonium ion and hydroxide ion to concentration of ammonia is ${\text{K}}_{\text{b}}$. By substituting the concentrations from equilibrium table, we can find the hydroxide ion concentration. The pOH value can be calculated by taking negative logarithm of concentration of hydroxide ion. Alternative way to find $\text{pH}$ value is Henderson-Hasselbalch equation. By substituting the ${\text{pK}}_{\text{a}}$ value and logarithm of concentrations of ammonium ion and ammonia, we can find the pH value.