Answer:
pH = 12.08.
Explanation:
Write the base reaction of methylamine:
[tex]\displaystyle \text{H$_2$O}_\text{($\ell$)} + \text{CH$_3$NH$_2$}_\text{(aq)} \rightleftharpoons \text{CH$_3$NH$_3$}^+_\text{ (aq)} + \text{OH}^-_\text{ (aq)}[/tex]
The equilibrium constant expression will hence be:
[tex]\displaystyle K_b = \frac{\left[\text{CH$_3$NH$_3$}^+\right]\left[\text{OH}^-\right]}{\left[\text{CH$_3$NH$_2$}\right]}[/tex]
As the reaction proceeds, x amounts of CH₃NH₃⁺ and OH⁻ will be formed and (0.35 M - x) amounts of CH₃NH₂ remains.
Assuming that the change to CH₃NH₂ is negligible, we have that:
[tex]\displaystyle \begin{aligned} K_b & = \frac{(x)(x)}{(0.35 -x)} \\ \\ (4.4\times 10^{-4}) & \approx \frac{x^2}{0.35} \\ \\ x & =0.012 \end{aligned}[/tex]
Hence, [OH⁻] = 0.012 M.
Find [H⁺]:
[tex]\displaystyle \begin{aligned} \ [\text{H}^+][\text{OH}^-] & = 1.0\times 10^{-14} \\ \\ [\text{H}^+] (0.012) & = 1.0\times 10^{-14} \\ \\ [\text{H}^+] & = 8.3\times 10^{-13} \text{ M}\end{aligned}[/tex]
Hence, the pH is:
[tex]\displaystyle \begin{aligned} \text{pH} & = -\log [\text{H}^+] \\ \\ & = -\log (8.3\times 10^{-13}) \\ \\ & = 12.08\end{aligned}[/tex]
In conclusion, the pH is about 12.08.
