Question

$H_{3}P{O}_4$ is a tribasic acid with $p{K}_{a_1},p{K}_{a_2}$ and $p{K}_{a_3}$ $2.12,7.21$ and $12.32$, respectively. It is used in fertiliser productions and its various salts are used in food, detergent, toothpaste, and in metal treatment.

Small quantities of $H_{3}P{O}_4$ are used in imparting the sour or tart taste to soft drinkes, such as Coca Cola, and beers, in which $H_{3}P{O}_4$ is present $0.05$% by weight $(density=1.0gm{L}^{-1})$. ${10}^{-3}M{H}_{3}P{O}_4(pH=7)$ is used in fertilisers as an aqueous soil digesting. Plants can absorb zinc in water soluble form only. Zinc phosphate is the source of zinc and $P{O}_2^{3-}$ ions in the soil. $K_{sp}$ of zonic phosphate

$=9.1\times {10}^{-33}$$\left[Z{n}^{2+}\right]$ ion in the soil is:

• A. $2.9\times {10}^{-11}M$
• B. $4.0\times {10}^{-10}M$
• C. $3.0\times {10}^{-6}M$
• D. $9.1\times {10}^{-5}M$

Answer 1

The correct option is A $2.9\times {10}^{-11}M$

$pH=7$
$\left[H_3{O}^{+}\right]={10}^{-7}$
pKa1, pKa2 and pKa3 are 2.12, 7.21 and 12.32 respectively. Hence, Ka1, Ka2 and Ka3 are $7.59\times {10}^{-3}$, $6.17\times {10}^{-8}$ and $4.79\times {10}^{-13}$ respectively.
Note: $K_a={10}^{-p{K}_a}$
$f_3=\frac{K_{a1}{K}_{a2}{K}_{a3}}{K_{a1}{K}_{a2}{K}_{a3}+K_{a1}{K}_{a2}\left[H_3{O}^{+}\right]+K_{a1}[H_3{O}^{+}{]}^2+[H_3{O}^{+}{]}^3}$
$f_3=\frac{7.59\times {10}^{-3}\times 6.17\times {10}^{-8}\times 4.79\times {10}^{-13}}{7.59\times {10}^{-3}\times 6.17\times {10}^{-8}\times 4.79\times {10}^{-13}+7.59\times {10}^{-3}\times 6.17\times {10}^{-8}\times {10}^{-7}+7.59\times {10}^{-3}\times ({10}^{-7}{)}^2+({10}^{-7}{)}^3}$
$f_3=1.83\times {10}^{-6}$
Let the solubility of $\left[Z{n}_3\right(P{O}_4{)}_2]$ be S M.
$\left[Z{n}^{2+}\right]=3S$ M
$\left[P{O}_4^{3-}\right]=f_3\times 2S$
$K_{sp}=\left[Z{n}^{2+}{]}^3\right[P{O}_4^{3-}{]}^2$
$9.1\times {10}^{-33}=(3S{)}^3\times (f_3\times 2S{)}^2$
$9.1\times {10}^{-33}=(3S{)}^3\times (1.83\times {10}^{-6}\times 2S{)}^2$
$S=3.0\times {10}^{-5}$ M
$\left[Z{n}^{2+}\right]=3S=3\times 3.0\times {10}^{-5}=9\times {10}^{-5}$ M
Hence, the option (D) is the answer.