Question

7. a) What is the concentration of Ba2+ in a saturated solution of BaF2 in pure water? Ksp (BaF2) = 2.45 x 10-5 ?

b) What is the concentration of Ba2+ in a saturated solution of BaF2 in a solution already containing
0.361 mol L-1 NaF?

Answer 1

a)

At equilibrium:

BaF2 <----> Ba2+ + 2 F-

   s 2s

Ksp = [Ba2+][F-]^2

2.45*10^-5=(s)*(2s)^2

2.45*10^-5= 4(s)^3

s = 1.83*10^-2 M

Answer: 1.83*10^-2 M

b)

NaF here is Strong electrolyte

It will dissociate completely to give [F-] = 0.361 M

At equilibrium:

BaF2 <----> Ba2+ + 2 F-

   s 0.361 + 2s

Ksp = [Ba2+][F-]^2

2.45*10^-5=(s)*(0.361+ 2s)^2

Since Ksp is small, s can be ignored as compared to 0.361

Above expression thus becomes:

2.45*10^-5=(s)*(0.361)^2

2.45*10^-5= (s) * 0.1303

s = 1.88*10^-4 M

Answer: 1.88*10^-4 M