It is of interest to decide if an analytical separation of the metal ions can be

Question

It is of interest to decide if an analytical separation of the metal ions can be effected by selective precipitation of carbonates from a solutiota 0.106 M in Ca2*. 0.106 M in Fe2+ and Feco Kp3.50*10 CaCO3Ksp 3Kp3.80x10 To analyze this problem, answer the following questions (1) What carbonate concentration is needed to precipitate 99.9% of the metal that forms the least soluble carbonate? (2) When 99.9% of the least soluble carbonate has precipitated, will all of the metal that forms the more soluble carbonate still remain in solutio (3) What is the upper limit on the carbonate ion concentration if the more soluble compound is not to precipitate? (4) If the [CO32] is at this upper limit, what percentage of the metal that forms the least soluble carbonate remains in solution? yes no 1%

Explanation / Answer

1) Since FeCO3 and CaCO3 form similar type of salts, hence, we compare the Ksp values to decide which salt is more soluble. Since CaCO3 has a higher Ksp value, hence, CaCO3 is more soluble. Consequently, FeCO3 is the least soluble carbonate.

We wish to precipitate 99.9% Fe2+ as FeCO3; therefore, the percentage of Fe2+ in solution = (100.0 – 99.9)% = 0.1%.

Therefore, concentration of Fe2+ left in solution = (0.1)%*(0.106 M) = (0.1/100)*(0.106 M) = 1.06*10-4 M.

Use this value for the concentration of Fe2+ to determine the concentration of CO32- required to precipitate 99.9% Fe2+. Use the Kspvalue.

Ksp = [Fe2+][CO32-]

====> 3.50*10-11 = (1.04*10-4)*[CO32-]

====> [CO32-] = (3.50*10-11)/(1.04*10-4) = 3.365*10-7 ? 3.36*10-7

The carbonate concentration required = 3.36*10-7 M (ans).

2) Use the Ksp for CaCO3 and the carbonate ion concentration from above to determine the concentration of Ca2+ when 99.9% Fe2+ has precipitated.

CaCO3 (s) <======> Ca2+ (aq) + CO32- (aq)

Ksp = [Ca2+][CO32-]

====> 3.80*10-9 = [Ca2+]*(3.36*10-7)

====> [Ca2+] = (3.80*10-9)/(3.36*10-7) = 0.0113 ? 0.012

The concentration of Ca2+ remaining in solution when 99.9% Fe2+ has precipitated is 0.011 M. We started with 0.106 M Ca2+ and we are left with 0.011 M Ca2+. Therefore, the percentage of Ca2+ remaining in solution = (0.011 M)/(0.106 M)*100% = 10.38%.

Thus, all of Ca2+ doesn’t remain in solution (ans).

3) Use the Ksp for CaCO3 to determine [CO32-] when CaCO3 just begins to precipitate.

Ksp = [Ca2+][CO32-]

====> 3.80*10-9 = (0.106)*[CO32-]

====> [CO32-] = (3.80*10-9)/(0.106) = 3.585*10-8 ? 3.58*10-8.

When the carbonate ion concentration is 3.58*10-8 M, CaCO3 begins to precipitate. Hence, in order to keep Ca2+ ions in solution, the carbonate ion concentration must be less than 3.58*10-8 M. The maximum carbonate ion concentration which will keep Ca2+ in solution is 3.57*10-8 M (ans).

4) Use the [CO32-] from the above step and the Ksp for FeCO3 to determine [Fe2+].

Ksp = [Fe2+][CO32-]

====> 3.50*10-11 = [Fe2+]*(3.57*10-8)

====> [Fe2+] = (3.50*10-11)/(3.57*10-8) = 9.804*10-4 ? 9.80*10-4

The [Fe2+] is 9.80*10-4 M and the percentage of Fe2+ = (9.80*10-4 M)/(0.106 M)*100% = 0.924% ? 0.92% (ans).

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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