Question

0.1M of HA is titrated with 0.1 M NaOH, calculate the pH at the end point. Given, Ka (HA)=$5 \times {10^{ - 6}}$ and a<<1.

Answer 1

Hint: In this question use the concept of equilibrium ${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}}$ to find the value of $\alpha $. Then subtract $\alpha $ from the total number of moles of substance to get the number of ions left un-neutralized. Find the log of the value obtained and subtract it from 14 to get the value of pH at the end point.

Complete answer:
Formula Used:
$HA \rightleftharpoons {H^ + } + {A^ - }$
${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}}$
According to question
c = 0.1 M of HA (given)
$HA \rightleftharpoons {H^ + } + {A^ - }$
At equilibrium
${K_a} = \dfrac{{[{H^ + }][{A^ - }]}}{{[HA]}} = \dfrac{{c\alpha \times c\alpha }}{{c(1 - \alpha )}}$
given $a < < 1,$then(1-a)$ \approx $1 can be neglected

$5 \times {10^{ - 6}} = \dfrac{{c{\alpha ^2}}}{1}$
$\alpha = 0.7071 \times {10^{ - 3}}$
$[{H^ + }] = [c\alpha ] = 0.707 \times {10^{ - 4}}M$
$0.7071 \times {10^{ - 3}}M$ will neutralize $0.7071 \times {10^{ - 4}}M$ of [OH-] of 0.1M NaOH
[OH-] left un-neutralized = 0.1M-$0.7071 \times {10^{ - 4}}M$ = $9.99 \times {10^{ - 2}}$M
pOH= -log[OH-] = -log[$9.99 \times {10^{ - 2}}$] =1.00
pH=14-1
pH=13.

The pH at the end point is 13.

Note: You must understand that titration is the method of chemical analysis where the quantity of any sample constituent is calculated by applying an exactly specified quantity of another component to the measured sample with which the target constituent responds in a definite and established proportion.