Answer:
9.58
Explanation:
Given that:
[tex]pK_{b}=8.75[/tex]
[tex]K_{b}=10^{-8.75}=1.78\times 10^{-9}[/tex]
Concentration = 0.215 M
Consider the ICE take for the dissociation of the base as:
C₅H₅N + H₂O ⇄ C₅H₅NH⁺ + OH⁻
At t=0 0.215 - -
At t =equilibrium (0.215-x) x x
The expression for dissociation constant is:
[tex]K_{b}=\frac {\left [ C_5H_5NH^{+} \right ]\left [ {OH}^- \right ]}{[C_5H_5N]}[/tex]
[tex]1.78\times 10^{-9}=\frac {x^2}{0.215-x}[/tex]
x is very small, so (0.215 - x) ≅ 0.215
Solving for x, we get:
x = 3.827×10⁻⁵ M
pOH = -log[OH⁻ ] = -log(3.827×10⁻⁵) = 4.42
pH = 14 - pOH = 14 - 4.42 = 9.58
