Question

2. A solution of K2SO4 is added to a solution that is 0.015 M in Ba^2+...

2. A solution of K2SO4 is added to a solution that is 0.015 M in Ba^2+ and 0.015 M in Sr^2+.

a. What is the necessary concentration of sulfate ion to begin precipitation?

b. Does Ba^+2 or Sr^+2 precipitate first? BaSO4: Ksp = 1.1 x 10^-10; SrSO4: Ksp = 3.2 x 10^-7

Homework Answers

Answer #1

BaSO4(s) ----------------------> Ba^2+ (aq) + SO4^2- (aq)

Ksp   = [Ba^2+][SO4^2-]

1.1*10^-10   = 0.015[SO4^2-]

[SO4^2-]    = 1.1*10^-10/0.015   = 7.3*10^-9 M

SrSO4(s) -------------------->Sr^2+ (aq) + SO4^2-

   Ksp   = [Sr^2+][SO4^2-]

3.2*10^-7   = 0.015[SO4^2-]

[SO4^2-]    = 3.2*10^-7/0.015   = 2.13*10^-5 M

there fore the precipitate will start at 7.3*10^-9 M

[SO4^2-] = 7.3*10^-9 M

b. Ba^2+ will form first precipitate because of less ksp value of BaSO4 than SrSO4

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