Question

Calculate the pH of a 0.20 M C₂H₅NH₂ solution (K_b = 5.6 × 10^-4).

Short answer

pH of a 0.20 M C₂H₅NH₂ solution is 12.02.

Step-by-step solution

Calculation of [OH-] of Ethylamine

Ethylamine reacts with water as shown

$$\begin{gathered} {\mathrm{C}}_2{\mathrm{H}}_5{{\mathrm{NH}}_2}_{\left(\mathrm{aq}\right)}+{\mathrm{H}}_2\mathrm{O}\iff {\mathrm{C}}_2{\mathrm{H}}_5{{\mathrm{NH}}^{+}}_{\left(\mathrm{aq}\right)}+{{\mathrm{OH}}^{-}}_{\left(\mathrm{aq}\right)} \\ 0.2\mathrm{\_}00 \\ 0.2-\mathrm{x}-\mathrm{x}\mathrm{x} \end{gathered}$$

[OH^-] ion concentration can be calculated as

localid="1650273206559" $$\begin{gathered} {\mathrm{K}}_{\mathrm{b}}=\frac{{\mathrm{x}}^2}{0.2-\mathrm{x}}=5.6\times {10}^{-4} \\ {\mathrm{x}}^2=5.6\times {10}^{-4}\times 0.2-5.6\times {10}^{-4}\mathrm{x} \\ {\mathrm{x}}^2+5.6\times {10}^{-4}\mathrm{x}-1.12\times {10}^{-4}=0 \end{gathered}$$

$$\begin{gathered} \mathrm{for}\mathrm{solving} \\ \frac{-\mathrm{b}\pm \sqrt{{\mathrm{b}}^2-4\mathrm{ac}}}{2\mathrm{a}}=\frac{-5.6\times {10}^{-4}\pm \sqrt{{(-5.6\times {10}^{-4})}^2-4\left(1\right)-(1.12\times {10}^{-4})}}{2\left(1\right)}\Rightarrow \mathrm{x}=1.06\times {10}^{-4} \end{gathered}$$

Calculating pH from [OH-]

[H⁺] ion concentration can be calculated as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1\times {10}^{-14} \\ \left[{\mathrm{H}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}=\frac{1\times {10}^{-14}}{1.06\times {10}^{-2}}=9.434\times {10}^{-13} \end{gathered}$$

Now pH can be calculated as

${\mathrm{p}}^{\mathrm{H}}=-\log \left[{\mathrm{H}}^{+}\right]=-\log (9.434\times {10}^{-13})=12.02$