Chapter 16, Problem 16.97QP

Calculate the pH of a 0.91 M C₂H₅NH₃I solution. (K_b for C₂H₅NH₂ = 5.6 × 10⁻⁴.)

Interpretation Introduction

Interpretation:

The pH of a $\text{0}\text{.91}M{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ solution has to be calculated

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{(aq)}+{\text{H}}_{\text{2}}\text{O }\to {\text{ H}}_{\text{3}}{\text{O}}^{+}{}_{(aq)}+{\text{A}}^{\text{-}}{}_{(aq)}$ is given below.

$K_a=\frac{[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$  is concentration of hydronium ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\to {\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between $K_a$ and $K_b$

$K_a\times K_b=K_w$

Where,

$K_w$ is autoionization of water.

$\begin{array}{l}\text{The autoionization of water}:\\ K_w=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ =1.0\times {10}^{-14}\end{array}$

Relationship between pH and pOH:

For calculation of pH, the following formula is used,

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

For calculation of pOH, the following formula is used,

$\text{pOH}\text{=}-\log {\text{[OH}}^{\text{-}}\text{]}$

From definition of pH and pOH, we get at room temperature,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14}$

To Calculate: The pH of a $\text{0}\text{.91}M{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ solution

Answer to Problem 16.97QP

Answer

The pH of a $\text{0}\text{.91}M{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ solution is 5.39

Explanation of Solution

Record the given datas

The concentration of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ is 0.91 M

The $K_b$ for ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ is $5.6\times {10}^{-4}$

The concentration  of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ and base ionization constant of  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$ is given. Based on the given datas, the pH of 0.91 M ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ has to be calculated

Construct equilibrium table

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ produces ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{}^{+}$ and ${\text{I}}^{\text{-}}$ ions. The iodide ion is the anion of the strong acid $\text{HI}$. Therefore, it will not hydrolyze.

The ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{}^{+}$ ion is the conjugate acid of the weak base ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$. It hydrolyzes to produce ${\text{H}}^{\text{+}}$ and the weak base ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$:

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{{}^{\text{+}}}_{\text{(aq)}}\text{+}{\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

In order to determine the pH, we must first determine the concentration of hydronium ion produced by the hydrolysis.

Express the equilibrium concentration of all species in terms of initial concentrations and a single unknown x

	${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_3{{}^{\text{+}}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_2{}_{(aq)}+{\text{H}}_{\text{3}}{\text{O}}^{+}{}_{(aq)}$
Initial $(M)$	0.91 $-x$ $0.91-x$	$0.00$	$0.00$
Change $(M)$		$+x$	$+x$
Equilibrium $(M)$		$x$	$x$

Calculation $K_a$ value of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{}^{+}$

Calculate $K_a$ value of  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{}^{+}$ from $K_b$ value of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}$. The relationship is given as follows,

$K_a\times K_b=K_w$

The above equation can be rewritten as,

$\begin{array}{l}{K}_a=\frac{K_w}{K_b}\\ =\frac{1.0\times {10}^{-14}}{5.6\times {10}^{-4}}\\ =1.8\times {10}^{-11}\end{array}$

Solve for $x$

Write the ionization constant expression in terms of the equilibrium concentrations.  Knowing the value of the equilibrium constant $K_a$ , solve for $x$. Then find out the pH  as follows,

$\begin{array}{l}{K}_a=\frac{{\text{[C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{2}}{\text{]H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}{}^{\text{+}}\text{]}}\\ 1.8\times {10}^{-11}=\frac{x^2}{0.91-x}\\ \text{Assuming}\text{(0}\text{.91-}x\text{)}\approx \text{0}\text{.91,}\text{then}\\ =\frac{x^2}{0.91}\\ x={\text{[H}}^{\text{+}}\text{]}\\ \text{=4}\text{.05}\times {\text{10}}^{\text{-6}}M\\ \text{pH}\text{=}\text{-log(4}\text{.05}\times {\text{10}}^{\text{-6}})\\ =5.39\end{array}$

Therefore, the pH of the given solution is 5.39

Conclusion

The pH of a $\text{0}\text{.91}M{\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{NH}}_{\text{3}}\text{I}$ solution was calculated.