Approximate pH of 0.01 , ,M ,NaHA is calculated by : ( K_ a_1 , =… - Vidyadip

MCQ

Approximate $pH$ of $0.01\,\,M\,NaHA$ is calculated by $: ({K_{{a_1}}}\, = \,{10^{ - 6}}$ and ${K_{{a_2}}}\, = \,{10^{ - 8}}$ are ionization constant of $H_2A$ )

A
$pH\, = 7 + \frac{{p{K_{{a_1}}}}}{2} + \frac{{\log \,C}}{2}\,$
B
$pH\, = 7 - \frac{{p{K_{{a_1}}}}}{2} - \frac{{\log \,C}}{2}\,$
✓
$pH\, = \frac{{p{K_{{a_1}}}\, + \,p{K_{{a_2}}}}}{2}\,$
D
$pH\, = \frac{{p{K_{{a_1}}}\, - \,p{K_{{a_2}}}}}{2}\,$

✓

Answer

Correct option: C.

$pH\, = \frac{{p{K_{{a_1}}}\, + \,p{K_{{a_2}}}}}{2}\,$

c
Given,

$k a_1=10^{-6}$

$k a_2=10^{-8}$

$NaHA$ is the ampheprotic salt for ampheprotec sait,

$p H=\frac{p k a_1+p k a_2}{2}$

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