Question

15 mL of 0.05 M $AgN{O}_3$ is mixed with 45 mL of 0.03 M $K_{2}Cr{O}_4$. Then, which of the following is correct regarding $A{g}_{2}Cr{O}_4$ . $K_{sp}$ of $A{g}_{2}Cr{O}_4=1.9\times {10}^{-12}.$

• A. Precipitation occurs
• B. Precipitation does not occur
• C. Can't be determined from given data
• D. None of these

Answer 1

The correct option is A Precipitation occurs

Final volume$=15+45=60mL=0.06L$

Using these volume finding out the final concentrations:
$c=\frac{mole}{volumeinL}$
$\left[A{g}^{+}\right]=\frac{0.015\times 0.05}{0.060}=1.25\times {10}^{-2}M$

$\left[Cr{O}_4^{-}\right]=\frac{0.045\times 0.03}{0.060}=2.25\times {10}^{-2}M$

Ionic product for $A{g}_{2}Cr{O}_4$
$A{g}_{2}Cr{O}_4\left(s\right)\rightleftharpoons 2A{g}^{+}\left(s\right)+Cr{O}_4^{-2}\left(s\right)$

Ionic product $=\left[A{g}^{+}{]}^2\right[Cr{O}_4^{-2}]$
$=(1.25\times {10}^{-2}{)}^2(2.25\times {10}^{-2})$
$=3.51\times {10}^{-6}$

Ionic product $>K_{sp}$
$\therefore$ precipitation occurs.