Question

One $d{m}^3$ solution containing $10^{-5}$ moles each of $C{l}^{-}$ ions and $Cr{O}_4^{-2}$ ions is treated with $10^{-4}$ mole of silver nitrate. Which one of the following observations is made?
$[K_{sp}A{g}_{2}Cr{O}_4=4\times 10^{-12}]$
$[K_{sp}AgCl=1\times 10^{-10}]$

• A. Silver chromate gets precipitated first.
• B. Precipitation does not occur.
• C. Both silver chromate and silver chloride start precipitating simultaneously.
• D. Silver chloride gets precipitated first.

Answer 1

Answer: (D)

For precipitation,

ionic product $>$ solubility product $\left(K_{sp}\right)$

For $A{g}_{2}Cr{O}_4$

ionic product $={\left[A{g}^{+}\right]}^2\left[Cr{O}_4^{-}\right]$

$={\left(10^{-4}\right)}^2\left(10^{-5}\right)=10^{-13}$

$K_{sp}$ of $A{g}_{2}Cr{O}_4=4\times 10^{-12}$

Here, $K_{\text{sp }}>IP$

Thus, no precipitate is obtained. For AgCl,

ionic product $=\left[A{g}^{+}\right]\left[C{l}^{-}\right]$

$=\left[10^{-4}\right]\left[10^{-5}\right]$

$=10^{-9}$

$K_{sp}(AgCl)=1\times 10^{-10}$

Here, $IP>K_{sp}$

So, precipitate will form.

Thus, silver chloride gets precipitated first.