Question

Sulphide ion in alkaline solution reacts with solid sulphur to form polysulphide ions having formula , $S_2^{2-},S_3^{2-},S_4^{2-}$, etc. if $K_1=12$ for $S+S^2\rightleftharpoons S_2^{2-}$ and $K_2=132$ for $2S+S^2\rightleftharpoons S_3^{2-}$, calculate $K_3$ for $S+S_2^{2-}\rightleftharpoons S_3^{2-}$ Give your answer by dividing $5.5$

Answer 1

Answer: 2.00

$S+S^2\rightleftharpoons S_2^{2-}{K}_1=12...(1)$
$2S+S^{2-}\rightleftharpoons S_3^{2-}{K}_2=132...(2)$
$S+S_2^{2-}\rightleftharpoons S_3^{2-}{K}_3=?$
$K_1=\frac{\left(S_2{}^{2-}\right)}{(S)\times \left(S^{2-}\right)}$
equation $(II÷I)$
$=\frac{K_2}{K_1}=\frac{\left(S_3{}^{2-}\right)}{(S{)}^2\times \left(S^{2-}\right)}\times \frac{(S)\left(S^{2-}\right)}{\left(S_2{}^{2-}\right)}$
$K_3=\frac{\left(S_3{}^{2-}\right)}{(S)\times \left(S_2^{2-}\right)}=\frac{K_2}{K_1}=\frac{\left(S_3{}^{2-}\right)}{(S)\times \left(S_2^{2-}\right)}$
$K_3=\frac{\left(S_3{}^{2-}\right)}{(S)\times \left(S_2{}^{2-}\right)}=\frac{132}{12}$
$=11$