JEE Main levelNumerical answerDifficulty 2 of 3ia082
Find the solubility of AgBr in 0.10MNaX2SX2OX3 solution, using AgBr(s)+2SX2OX3X2−⇌[Ag(SX2OX3)X2]X3−+BrX− with Ksp(AgBr)=5.0×10−13 and Kf([Ag(SX2OX3)X2]X3−)=2.9×1013. Give the answer in units of 10−2molL−1 (two decimals).
Numerical answer type, answer in x 10^-2 mol/L, tolerance 0.01.
Answer: 4.42 x 10^-2 mol/L
K=KspKf=14.5. With s=[Ag(SX2OX3)X2X3−]=[BrX−] and [SX2OX3X2−]=0.10−2s: s/(0.10−2s)=K=3.808, so s=0.10×3.808/(1+2×3.808)=0.0442M, i.e. 4.42 in units of 10−2M.