A solution is 0.020 M in barium chloride, BaCI_2, and 0.010 M in strontium chlor

Question

A solution is 0.020 M in barium chloride, BaCI_2, and 0.010 M in strontium chloride. SrCI_2. A concentrated sodium sulfate. Na_2SO_4, solution is added to the mixture. (Assume that the Na_2SO_4 solution is so concentrated that the volume change in the Ba-Sr solution can be neglected.) Which ion will precipitate first? Show work to justify your choice. No credit will be given for guesses. When the second ion just begins to precipitate, what is the residual concentration of the first ion? What fraction of the original amount of the first ion is left in solution?

Explanation / Answer

A.

Ksp of BaSO4 = 1.1*10-10

Ksp of SrSO4 = 3.4*10-7

in order to determine which one will be precipited first, the concentration of SO42- has to calculate...

for, BaSO4,..... [  SO42- ] = 1.1*10-10 / 0.020 = 55*10-10 M

for, SrSO4,..... [  SO42- ] = 3.4*10-7 / 0.010 = 34*10-6 M

so, the concentration of sulphate is very much less for BaSO4 then SrSO4, hence, BaSO4 will be precipitate first.

B. when the SrSO4 will be precipitate, the concentration of sulphate will be 34*10-6 M. To calculate the residual concentration of BaSO4 at the time of precipitation of SrSO4, the value of sulphate, 34*10-6 M has to put in the Ksp expression of BaSO4.

hence, the concentration of first ion will be

= 1.1*10-10 / 34*10-6 =0.032 *10-4 M

C. fraction of first ion will be = (0.032 *10-4 M / 0.020 M) * 100 % = 0.016 %

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.