The KspofAg2CrO4is1.1×10−12at298K. The solubility (in mol/L) of Ag2CrO4 in a 0.1MAgNO3 solution is
A
1.1×10−11
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1.1×10−10
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
1.1×10−12
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1.1×10−9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B1.1×10−10 PLAN In presence common ion (in this case Ag+ ion) solubiility of sparingly soluble salt is decreased.
Let solubility of Ag2CrO4 in presence of 0.1 M AgNO3=x Ag2CrO4⇌2Ag++CrO2−4 AgNO3⇌Ag+0.1+NO−30.1 Total [Ag+] = (2x + 0.1) M ≈ 0.1 M as x <<< 0.1 M [CrO2−4]=xM Thus, [Ag+]2[CrO2−4]=Ksp (0.1)2(x)=1.1×1012 ∵x=1.1×10−10M