Question

The pH of a 0.15 M butylamine, C&H3NH2 solution is 12.0 at 25°C. Calculate the dissociation

Answer 1

The dissociation  constant of the base : 7.4 x 10⁻⁴

Further explanation

Butylamine, C4H9NH2 Is A Weak Base

Kb is the dissociation  constant of the base.​

LOH (aq) ---> L⁺ (aq) + OH⁻ (aq)  

$\rm Kb=\dfrac{[L][OH^-]}{[LOH]}$

[OH⁻] for weak base can be formulated :

$\tt [OH^-]=\sqrt{Kb.M}$

pH of solution : 12

pH+pOH=14, so pOH :

14-12 = 2, then :

$\tt [OH^-]=10^{-pOH}\\\\(OH^-]=10^{-2}$

the the dissociation  constant (Kb) =

$\tt 10^{-2}=\sqrt{Kb.0.15}\\\\10^{-4}=Kb\times 0.15\\\\Kb=\dfrac{10^{-4}}{0.15}=6.6\times 10^{-4}$

Or you can use from ICE method :

C4H9NH2(aq) + H2O(l) ⇌ C4H9NH3+(aq) + OH-(aq)

0.15

x                                                x                        x

0.15-x                                        x                        x

$\tt Kb=\dfrac{x^2}{0.15-x}\rightarrow x=[OH^-]\\\\Kb=\dfrac{10^{-4}}{0.15-10^{-2}}=7.14\times 10^{-4}$