Question

. Calculate the pPb2+ in a solution obtained by titrating 25.0 mL of 0.088 M NaI with a) 22.0 mL b) Ve c) 28.0 mL of 0.046 M of Pb(NO3)2 . Ksp = 7.9 x 10-9

Answer 1

a)

moles of NaI = 25 x 0.088 / 1000 = 2.2 x 10^-3

moles of Pb+2 = 22 x 0.046 / 1000 = 1.01 x 10^-3

Pb+2       + 2I-    -----------------> PbI2

1                2                                  1

2.2 x 10^-3    1.01 x 10^-3                    

here limiting reagent is I- .

remaining Pb+2 = 2.2 x 10^-3 - (1.01 x 10^-3 / 2) = 1.694 x 10^-3 mol

concentration of Pb+2 = 1.694 x 10^-3 / 0.025 + 0.022 = 0.036 M

pPb+2 = 1.44

b)

PbI2   ------------> Pb+2 +   2I-

Ksp = [Pb+2][I-]^2

7.9 x 10^-9 = (S)(2S)^2

S = 1.26 x 10^-3

pPb+2 = 2.90

c)

concentration of Pb+2 = 28 x 0.046 / 28 + 25 = 0.0243 M

concentration of I- = 25 x 0.088 / 28 + 25 = 0.0415

limiting reagent is I-.

remaining Pb+2 = 0.0243 - 0.0415 / 2 = 3.55 x 10^-3 M

pPb+2 = 2.45