Question

Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylamine is 5.0× 10–4.

Answer 1

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of $[OH^-]$ be x

Initial concentration of $[CH3NH_2]$, c = 0.230 M

        $CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-$

at eq'm  c-x                         x                    x

Expression of $K_b$:

$K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}$

Since ,methyl-amine is a weak base,c>>x so $c-x\approx c$.

$K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}$

Solving for x, we get:

$x=1.07\times 10^{-2} M$

Given, HCl with 0.230 M , it dissociates fully in water which means $[H^+]$ = 0.230 M

$[OH^-]=[H^+]$ will result in neutral solution, since $[OH^-]$

Remaining $[H^+]$ after neutralizing $[OH^-]$ions

$[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M$

$pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65$

The pH at the equivalence point for the titration will be 0.65.