Question

The equilibrium constant for the ionization of $R{NH}_2\left(g\right)$ in water as
$R{NH}_2\left(g\right)+H_{2}O\left(l\right)\rightleftharpoons R{{NH}_3}^{+}\left(aq\right)+{OH}^{-}\left(aq\right)$
is ${8\times 10}^{-6}$ at ${25}^{o}C$. Find the pH of a solution at equilibrium when pressure of $R{NH}_2\left(g\right)$ is $0.5$ bar:

• A. $\approx 12.3$
• B. $\approx 11.3$
• C. $\approx 11.45$
• D. None of the above

Answer 1

Answer: (B)

$\approx 11.3$

For this reaction:

$R{NH}_2\left(g\right)+H_{2}O\left(l\right)\rightleftharpoons {\underset{x}{R{NH}_3}}^{+}\left(aq\right)+\underset{x}{{OH}^{-}\left(aq\right)}$

Given, $p_{RN{H}_2}=0.5$ bar

$K_p=\frac{\left[O{H}^{-}\right]\left[RN{H}_3^{+}\right]}{\left[RN{H}_2\right]}=\frac{x\times x}{0.5}=8\times {10}^{-6};$

$\Rightarrow x=\sqrt{0.5\times 8\times {10}^{-6}}=2\times {10}^{-3}$

$pOH=-log[OH{]}^{-}=2.7;$

$So,pH=14-pOH=14-2.7=\mathrm{11.3.}$

Option B is correct.