Answer:
Option B
Explanation:
Plan in presence of common ion (in this case Ag+ ion ) solubility of sparingly soluble salt is decreased
Let solubility of Ag2CrO4 in presence of 0.1 m
AgNO3 =x
$Ag_{2}CrO_{4}\rightleftharpoons 2Ag^{+}+CrO_2^{4-}$
2x x
$ AgNO_{3}\rightleftharpoons Ag^{+}+NO_3^{-}$
0.1 0.1
Total [Ag+]= [2x+0.1]M
= 0.1 M as x<<< 0.1 M
$ [Cr_{4}^{2-}]=x M$
thus, $[Ag^{+}]^{2+}[CrO^{2-}_{4}]= K_{sp}$
(0.1)2 (x)=1.1 x 10-12
$\therefore$ x= $1.1\times 10^{-10}$ M
