Answer:
5.65
Explanation:
Given that:
[tex]pK_{b}\ of\ CH_3NH_2=3.44[/tex]
[tex]K_{b}\ of\ CH_3NH_2=10^{-3.44}=3.6308\times 10^{-4}[/tex]
[tex]K_a\ of\ CH_3NH_3^+Cl^-=\frac {K_w}{K_b}=\frac {10^{-14}}{3.6308\times 10^{-4}}=2.7542\times 10^{-11}[/tex]
Concentration = 0.18 M
Consider the ICE take for the dissociation as:
[tex]CH_3NH_3^+[/tex] ⇄ H⁺ + [tex]CH_3NH_2[/tex]
At t=0 0.18 - -
At t =equilibrium (0.18-x) x x
The expression for dissociation constant of acetic acid is:
[tex]K_{a}=\frac {\left [ H^{+} \right ]\left [CH_3NH_2 \right ]}{[CH_3NH_3^+]}[/tex]
[tex]2.7542\times 10^{-11}=\frac {x^2}{0.18-x}[/tex]
x is very small, so (0.18 - x) ≅ 0.18
Solving for x, we get:
x = 0.2227×10⁻⁵ M
pH = -log[H⁺] = -log(0.2227×10⁻⁵) = 5.65
