Question

0.1 m of ha is titrated with 0.1 m naoh, calculate the ph at end point. Given ka (ha) = 5 Ă— 10 –6 and ďˇ << 1

Answer 1

Answer:The pH at end point is 13.

Explanation:

c = 0.1 M of HA

         $HA\rightleftharpoons H^++A^-$

At eq'm  c-c$\alpha$  c$\alpha$ c$\alpha$

$K_a=\frac{[H^+][A^-]}{[HA]}=\frac{c\alpha \times c\alpha}{c(1-\alpha )}$

given $\alpha <<1, then (1-\alpha )\approx 1$ can be neglected

$5\times10^{-6}=\frac{c\alpha ^2}{1}$

$\alpha =0.7071\times 10^{-3}$

$[H^+]=[c\alpha]=0.707\times 10^{-4} M$

$0.707\times 10^{-4} M$ will neutralize $0.707\times 10^{-4} M$ of $[OH^-]$ of 0.1 M NaOH

$[OH^-]$ left unneutralized = $0.1 M-0.707\times 10^{-4} M=9.99\times 10^{-2}M$

$pOH=-\log[OH^-]=-\log[9.99\times 10^{-2}]=1.00$

pH=14-pOH=14-1.00=13

The pH at end point is 13.