Question

One $dm^3$ solution containing $10^{-5}$ moles each of $Cl^-$ ions and $CrO_4^{-2}$ ions is treated with $10^{-4}$ mole of silver nitrate. Which one of the following observations is made?
$[K_{sp }Ag_2CrO_4 =4\times10^{-12}]$
$[K_{sp} AgCl=1\times10^{-10}]$

• 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 $\left(K_{s p}\right)$

For $Ag _{2} CrO _{4}$

ionic product $=\left[ Ag ^{+}\right]^{2}\left[ CrO _{4}^{-}\right]$

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

$K_{ sp }$ of $Ag _{2} CrO _{4}=4 \times 10^{-12}$

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

Thus, no precipitate is obtained. For AgCl,

ionic product $=\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]$

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

$=10^{-9}$

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

Here, $IP >K_{ sp }$

So, precipitate will form.

Thus, silver chloride gets precipitated first.