Question

The solubility product constant of $A{g}_{2}Cr{O}_4$ and $AgBr$ are $1.1\times 10^{-12}$ and $5.0\times 10^{-13}$ respectively. The ratio of the molarities is

• A. $9.4$
• B. $100$
• C. $91.94$
• D. $85$

Answer 1

Answer: (C)

$K_{sp}\left(A{g}_{2}Cr{O}_4\right)>K_{sp}(AgBr)$
$\therefore A{g}_{2}Cr{O}_4$ is more soluble
$A{g}_{2}Cr{O}_4(s)\rightleftharpoons 2A{g}_{(aq)}^{+}+Cr{O}_4^{-2}(aq)$,
$K_{sp}=4S^{3}and\therefore S={\left(\frac{1.1\times 10^{-12}}{4}\right)}^{13}=6.50\times 10^{-5}M$
$AgB{r}_{\left(s\right)}\rightleftharpoons A{g}_{aq}^{+}+B{r}_{aq}^{-};K_{sp}=S^2$
$\therefore S^{{}^{\prime }}={\left(5\times 10^{-13}\right)}^{12}=7.07\times 10^{-7}M$
The ratio o f the molarities $=\frac{S}{S^{{}^{\prime }}}=\frac{6.50\times 10^{-5}}{7.07\times 10^{-7}}=91.94$