Barium and strontium ions can be precipitated as sulfate salts. BaSO_4(s) Ba^2+(

Question

Barium and strontium ions can be precipitated as sulfate salts. BaSO_4(s) Ba^2+(aq) + SO_4^2-(aq) K_sp = 1.1 times 10^-10 SrSO_4(s) Sr^2+(aq) + SO_4^2-(aq) K_sp = 3.4 times 10^-7 BaCl_2 = 208.2341 g/mol BaSO_4 = 233.3909 g/mol SrCl_2 = 158.5274 g/mol SrSO_4 = 183.6842 g/mol a. Consider a solution made from a mixture of Bacl_2 and SrCl_2. In the solution, Ba^2+ and Sr^2+ each have a concentration of 0.100 M. If sodium sulfate is added to the solution, which sulfate precipitates first? Explain. b. What will be the concentration (mol/L) of the first ion that precipitates when the second ion (the more soluble ion) starts to precipitate? Calculate the concentration of phosphate ion when silver phosphate starts to precipitate from a solution that is 0.0125 M in Ag^+. Silver chromate (Ag_2CrO_4) has a K_sp of 9.0 times 10^-12 at 25.0 degree C. Predict the relative solubility (most soluble to least soluble) of silver chromate in the following solutions: I. pure water II 0.1 M AgNO_3 Ill. 0.1 M Na_2CrO_4

Explanation / Answer

(20)

(a) Since Ksp of BaSO4 < SrSO4, hence BaSO4 gets precipitated first.

(b)

SrSO4 (s) = Sr2+ (aq.) + SO42- (aq.)

Ksp = [Sr2+][SO42-]

3.4 * 10-7 = (0.100)[SO42-]

[SO42-] = 3.4 * 10-6M, This is the concentration of sulphate ion when second ion Sr2+ gets precipitated.

Now,

BaSO4 (s) = Ba2+ (aq.) + SO42- (aq.)

Ksp = [Ba2+][SO42-]

1.1 * 10-10 = [Ba2+](3.4*10-6)

[Ba2+] = 3.24 * 10-5 M

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