Question

A solution is ${10}^{-4}MinC{l}^{-},{10}^{-5}MinB{r}^{-},{10}^{-3}Min{I}^{-}.AgN{O}_3\left(s\right)$ is added slowly to the solution. The minimum concentration of $A{g}^{+}$ required to start precipitation of all three ions is ${10}^{-x}$ then the value of x is:
[Given , $K_{sp\left(AgCl\right)}={10}^{-10},K_{sp\left(AgBr\right)}={10}^{-13},K_{sp\left(AgI\right)}={10}^{-17}]$

• A. 8
• B. 14
• C. 5
• D. 6

Answer 1

The correct option is D 6

$\left[A{g}^{+}\right]=\frac{{10}^{-10}}{{10}^{-4}}={10}^{-6}M$ for $AgClppt.$

$\left[A{g}^{+}\right]=\frac{{10}^{-13}}{{10}^{-5}}={10}^{-8}M$ for $AgBrppt.$

$\left[A{g}^{+}\right]=\frac{{10}^{-17}}{{10}^{-3}}={10}^{-14}M$ for $AgIppt.$

So the minimum concentration required is ${10}^{-6}M$.

So the value of $x$ is 6.