IIT JEE Advanced 2006 Paper 1

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) =

kk(Sm−1)

C(molm−3)

⇒ k = λm×C
Therefore,
kAg+ = (6×10−3)(4×10−4) = 24 × 10−7Sm−1
KBr+ = (8×10−3)(3×10−4) = 24 × 10−7Sm−1
kNO3− = (7×10−3)×(10−4) = 7 × 10−7Sm−1
Ktotal = (24 + 24 + 7) × 10−7Sm−1 = 55 × 10−7Sm−1