JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia209
AgX(s)+2NHX3[Ag(NHX3)X2]X++XX−. Given Kf([Ag(NHX3)X2]+)=1.6×107; Ksp: AgCl=1.6×10−10, AgBr=5.0×10−13, AgI=8.5×10−17. Treat the free [NHX3] as constant at 1.0M. Select the correct statement(s).
A0.010mol of AgCl dissolves completely in 1.0L of such ammonia
B0.010mol of AgBr dissolves completely in 1.0L of such ammonia
CThe equilibrium constant for dissolution of AgBr in ammonia is 8×10−6
DAgI is more soluble than AgBr in ammonia because its Ksp is smaller
Answer: A, C
For each salt s=KspKf[NHX3]: AgCl 0.0506, AgBr 0.00283, AgI3.7×10−5M. So 0.010M of AgCl dissolves (0.0506>0.010) but AgBr does not (0.00283<0.010). K for AgBr is 5.0×10−13×1.6×107=8.0×10−6. A smaller Ksp means lower solubility, so AgI is less soluble.