Question

CALCULATE PH OF SATURATED LEAD (II) FLOURIDE IN WATER

Answer 1

PbF2(S) <====> Pb2+(aq) + 2F-(aq)

Ksp = S*(2S)^2

s = solubility = (ksp/4)^(1/3)

   = (ksp/4)^(1/3)

      = (3.6 x 10^(-8) /4)^(1/3)

= 0.00208 M

concentration of F- = 2*0.00208 = 0.00416 M

F- is strong base. so that it gets hydrolysis

           F-(aq) +     H2O(l)   ---> HF(aq) + OH-(aq)
Initial       0.00416            0         0

equilibrium   0.00416 -x            x     x

Kb = [OH-][HF]/[F-] = 1.48*10^-11

(1.48*10^(-11)) = x^2 / (0.00416 -x)

x = 2.48*10^-7 M

Concentration of OH- = 2.48*10^-7 M

pOH = -log(2.48*10^(-7))

     = 6.6

pH = 14-6.6 = 7.4