3. A solution is 0.020 M in barium chloride, BaCl2, and 0.010 M in strontium chl

Question

3. A solution is 0.020 M in barium chloride, BaCl2, and 0.010 M in strontium chloride, 2. A concentrated sodium sulfate, Na2SO4, solution is added to the mixture. SrCl (Assume that the Na2so4 solution is so concentrated that the volume change in the Ba-Sr solution can be neglected.) (a) Which ion will precipitate first? Show work to justify your choice. No credit will be given for guesses. (b) When the second ion just begins to precipitate, what is the residual concentration of the first ion? (c) What fraction of the original amount of the first ion is left in solution?

Explanation / Answer

3. Precipitation reaction

(a) Of the two Ba2+ will precipitate first from solution as BaSO4. The Ksp of BaSO4 is 1.1 x 10^-10 is lower than the Ksp of SrSO4 3.4 x 10^-7, thus BaSO4 has lower solubility in solution as compared to SrSO4 and it precipitates out first from the solution while SrSO4 remained soluble.

(b) Concentration of [SO4^2-] required to precipitate the second ion Sr2+,

[SO4^2-] = Ksp/[Sr2+]

               = 3.4 x 10^-7/0.01

               = 3.4 x 10^-5 M

Now, calculate concentration of [Ba2+] still in solution at this [SO4^2-] concentration,

[Ba2+] = Ksp/[SO4^2-]

           = 1.1 x 10^-10/3.4 x 10^-5

           = 3.23 x 10^-6 M

So the residual concentration of first ion [Ba2+] when the second ion [Sr2+] begins to precipitate is 3.23 x 10^-6 M

(c) Fraction of the first ion remained in solution

= 3.23 x 10^-6 x 100/0.02

= 0.01615%

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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