Question

Given that at 298K $K_{sp}$ for AgCNS and AgBr are $1.0\times {10}^{-12}$ and $5.0\times {10}^{-13}$ respectively. This data alone is sufficient to calculate the simultaneous solubility of AgCNS and AgBr - True or False?

• A. True
• B. False

Answer 1

The correct option is A

True

First, let us write down the reactions:

$AgCNS\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+CN{S}^{-}\left(aq\right)$

It is a binary salt - $AgCNS\left(s\right)$. The same can be said about $AgBr\left(s\right)$

$AgBr\left(s\right)\rightleftharpoons A{g}^{+}\left(aq\right)+B{r}^{-}\left(aq\right)$

It is given that these two, otherwise separate equilibria are existing simultaneously. We see that there is a common ion. But here the common ion effect will not be applicable. Why? Technically, there are just three ions and two solid in solution. That s a better approach to take.

If the solubility of $AgCNS\left(s\right)$ is $x$ and the solubility of $AgCNS\left(s\right)$ is $y$, we have

Net $\left[A{g}^{+}\right]$ = $x+y$

$\left[CN{S}^{-}\right]$ = $x$ and $\left[B{r}^{-}\right]$ = $y$

$K_{sp}\left(AgSCN\right)$ = $\left[A{g}^{+}\right]\left[CN{S}^{-}\right]$ = $(x+y)x$ ---------- (1)

& $K_{sp}\left(AgBr\right)$ = $\left[A{g}^{+}\right]\left[B{r}^{-}\right]$ = $(x+y)y$ ------------(2)

We have been given the values of $K_{sp}\left(AgSCN\right)$ and $K_{sp}\left(AgBr\right)$

So we have two equations and two variables. But it would be simpler to divide (1) by (2) and eliminate (x+y) here to expedite calculation.

Dividing equation (1) by equation (2), we get

$\frac{x}{y}$ = $2$

Now we can eliminate x in equation (1) and get a quadratic in $y$

$(2y+y)2y$ = ${10}^{-12}$

Solving $y$ = $4.08\times {10}^{-7}M$

Thus Solubility of $AgBr$ = $4.08\times {10}^{-7}M$ {y > 0}

Substituting the value of $y$ = $4.08\times {10}^{-7}M$ in (1), and solving the quadratic in $x$,

We get $x$ = solubility of AgSCN = $8.16\times {10}^{-7}M$ { only positive value} So the answer is true. But we don't need to calculate the solubilities .