Question

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

Answer 1

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