Question

Sulphide ion reacts with solid sulphur
$S_{\left(aq\right)}^{2-}+S_{\left(s\right)}\rightleftharpoons S_{2\left(aq\right)}^{2-};K_1=10$
$S_{\left(aq\right)}^{2-}+2S_{\left(s\right)}\rightleftharpoons S_{3\left(aq\right)}^{2-};K_1^{\prime }=130$
The equilibrium constant for the formation of $S_3^{2-}\left(aq\right)$ from $S_2^{2-}\left(aq\right)$ and sulphur is:

• A. $10$
• B. $13$
• C. $130$
• D. $1300$

Answer 1

The correct option is B $13$

Given,

$S^{2-}+S\to S_2^{2-}$ $K_1=10$

$K_1=\frac{\left[S_2^{2-}\right]}{\left[S^{2-}\right]}$

$10=\frac{\left[S_2^{2-}\right]}{\left[S^{2-}\right]}..........\left(i\right)$

$S^{2-}+2S\to S_3^{2-}$ $K_1^{\prime }=130$

$K_1^{\prime }=\frac{\left[S_3^{2-}\right]}{\left[S_2^{2-}\right]}$

$130=\frac{\left[S_3^{2-}\right]}{\left[S^{2-}\right]}..........\left(ii\right)$

$S_2^{2-}+S\to S_3^{2-}$................(iii) $K^{\prime \prime }$

$K^{\prime \prime }=K_1^{\prime }/K_1$

Dividing (ii) by (i), we get

$K^{\prime \prime }=\frac{\left[S_3^{2-}\right]}{\left[S_2^{2-}\right]}=\frac{130}{10}=13$