Question

For $M{K}_{3}A$ (potassium salt of a tribasic acid $H_{3}A$) solution:

(Dissociation constants of acid are $k_{a_1}{k}_{a_2}\&k_{a_3};h<<1$)

• A. $pH=\frac{1}{2}(p{K}_w+p{K}_{a_3}-logc)$
• B. $pH=\frac{1}{2}(p{K}_w+p{K}_{a_2}+logc)$
• C. $pH=\frac{1}{2}(p{K}_w+p{K}_{a_1}+logc)$
• D. $pH=\frac{1}{2}(p{K}_w-p{K}_{a_3}-logc)$

Answer 1

Answer: (A)

$pH=\frac{1}{2}(p{K}_w+p{K}_{a_3}-logc)$

$K_{3}A$ i.e. $A^{3-}$

$A^{3-}+H_{2}O\rightleftharpoons H{A}^{2-}+O{H}^{-}{K}_3\frac{K_w}{K{a}_3}$

$H{A}^{2-}+H_{2}O\rightleftharpoons H_2{A}^{-}+O{H}^{-}{K}_2\frac{K_w}{K{a}^2}$

$H_2{A}^{-}+H_{2}O\rightleftharpoons H_{3}A+O{H}^{-}.K_1\frac{K_w}{K{a}_1}$

$K{a}_1>K{a}_2>K{a}_3$

$\therefore K_3>K_2>K_1$

$\therefore \left[O{H}^{-}\right]$ will be most form 1st reaction

$A^{3-}+H_{2}O\rightleftharpoons H{A}^{2-}+O{H}^{-}$

c o o

$c(1-\alpha )$ $c\alpha$ $c\alpha$

$\frac{c{\alpha }^2}{(1-\alpha )}=\frac{k_w}{K{a}_3}$

$\alpha$ is way less than $1$

$\therefore c{\alpha }^2\frac{K_w}{K{a}_3}$

$\Rightarrow \alpha =\sqrt{\frac{K_w}{cK{a}_3}}$

$\left[O{H}^{-}\right]=c\alpha$

$=\sqrt{\frac{c{K}_w}{K{a}_3}}$

$[H^{+}\frac{K_w}{\left[O{H}^{-}\right]}=\sqrt{\frac{K{a}_3{K}_w}{c}}$

$PH=\frac{1}{\alpha }(p{K}_w+pK{a}_3-\log C)$