Question

Be sure to answer all parts. Find the molar solubility of BaCro4 (Ksp=2.1 x 10-10) in (a) pure water * 10 M (b) 8.6 x 10-3 M Na Cr04 * 10M

Answer 1

a)

At equilibrium:

BaCrO4 <----> Ba2+ + CrO42-

   s s

Ksp = [Ba2+][CrO42-]

2.1*10^-10=(s)*(s)

2.1*10^-10= 1(s)^2

s = 1.449*10^-5 M

Answer: 1.4*10^-5 M

2)

Na2CrO4 here is Strong electrolyte

It will dissociate completely to give [CrO42-] = 0.0086 M

At equilibrium:

BaCrO4 <----> Ba2+ + CrO42-

   s 8.6*10^-3 + s

Ksp = [Ba2+][CrO42-]

2.1*10^-10=(s)*(8.6*10^-3+ s)

Since Ksp is small, s can be ignored as compared to 8.6*10^-3

Above expression thus becomes:

2.1*10^-10=(s)*(8.6*10^-3)

2.1*10^-10= (s) * 8.6*10^-3

s = 2.442*10^-8 M

Answer: 2.4*10^-8 M