Let the solubility of AgBr in presence of 10−7 M AgNO3 solution be S. AgBr →
Ag+
S
+
Br−
S
AgNO3 →
Ag+
10−7mol
+
NO3−
10−7mol
Then, [Ag+] = [S+10−3] m ; [Br] = S Ksp = [Ag+][Br−] So, 12 × 10−14 = (5 + 10−7) S S2+10−7S + 12 × 10−14 = 0 S = - 10−7±√10−14+4×12×10−14 (By ignoring negative root), we get S =
−10−7+7×10−7
2
= 3 × 10−7 m Now, substituting values, we get the concentration of ions as [Ag+] = S + 10−7 = 3 × 10−7+10−7 = 4 × 10−7 m = 4 × 10−4molm−3 [Br+] = 3 × 10−7 m = 3 × 10−4molm−3 [NO3−] = 10−7 m = 10−14molm−1 We know that λm(Sm2mol−1) =