Problem 17.74 with feedback A solution of Na2SO4 is added dropwise to a solution

Question

Problem 17.74 with feedback

A solution of Na2SO4 is added dropwise to a solution that is 1.2×102M in Ba2+ and 1.2×102M in Sr2+.

The solubility-product constants are as follows:

BaSO4:SrSO4:KspKsp==1.1×10103.2×107

Part A

Which cation will precipitate first?

SubmitMy AnswersGive Up

Correct

The salt that has the lower solubility-product constant, BaSO4, will precipitate first because it represents much lower concentrations for the cation and SO42.

Part B

What concentration of SO42 is necessary to begin precipitation? (Neglect volume changes.)

Express the molarity to two significant digits.

1.04•106

SubmitMy AnswersGive Up

Incorrect; Try Again

Part C

At what concentration of SO42 will the second cation begin to precipitate?

Express your answer using two significant figures.

1.04•105

SubmitMy AnswersGive Up

Incorrect; Try Again

Problem 17.74 with feedback

A solution of Na2SO4 is added dropwise to a solution that is 1.2×102M in Ba2+ and 1.2×102M in Sr2+.

The solubility-product constants are as follows:

BaSO4:SrSO4:KspKsp==1.1×10103.2×107

You may want to reference ( pages 759 - 762) Section 17.6 while completing this problem.

Part A

Which cation will precipitate first?

Ba2+ precipitates first. Sr2+ precipitates first.

SubmitMy AnswersGive Up

Correct

The salt that has the lower solubility-product constant, BaSO4, will precipitate first because it represents much lower concentrations for the cation and SO42.

Part B

What concentration of SO42 is necessary to begin precipitation? (Neglect volume changes.)

Express the molarity to two significant digits.

1.04•106

  M  

SubmitMy AnswersGive Up

Incorrect; Try Again

Part C

At what concentration of SO42 will the second cation begin to precipitate?

Express your answer using two significant figures.

1.04•105

  M  

SubmitMy AnswersGive Up

Incorrect; Try Again

Explanation / Answer

b) Ksp = [Ba2+] [SO42-]
1.1x10-10 = [0.0120 M] [SO4--]; solving, [SO4--] = 9.166 x10-9 M.
Ksp = [Sr2+] [SO42-]
3.2x10-7 = [0.012 M] [SO42-] ; solving, [SO42-] = 2.66 x10-5 M.
Therefore, precipitation will start to occur at [SO42-] = 9.166x10-9 M.

c) 2.66 x10-5 M

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---