Be sure to answer all parts. find the molar solubility of bacro4 (ksp = 2.1 * 10-10) in (a) pure water * 10 m (b) 1.6 * 10-3 m na2cro4 * 10 m

A) in pure water:

by using ICE table:

According to the reaction equation:

BaCrO4 (s) → Ba^2 + (aq) + CrO4^2 - (aq)

initial 0 0

change + X + X

Equ X X

when Ksp = [Ba^2+][CrO4^2-]

by substitution:

2.1 x 10^-10 = X * X

∴X = √2.1 x 10*-10

∴X = 1.4 x 10^-5

∴ the solubility = X = 1.4 X 10^-5

B) In 1.6 x 10^-3 m Na2CrO4

by using ICE table:

According to the reaction equation:

BaCrO4 (s) → Ba^2 + (aq) + CrO4^2 - (aq)

initial 0 0.0016

Change + X + X

Equ X X+0.0016

when Ksp = [Ba^2+][CrO4^2-]

by substitution:

2.1 x 10^-10 = X * (X+0.0016) by solving for X

∴ X = 1.3 x 10^-7

∴ solubility = X = 1.3 x 10^-7