12. Selective precipitation of Ba2+ and Pb2+ with CrO4^2- addition, (C) PbCrO4 w
ID: 699846 • Letter: 1
Question
12. Selective precipitation of Ba2+ and Pb2+ with CrO4^2- addition,
(C) PbCrO4 will precipitate first and at [CrO4^2-] = 8.1 x 10^-10 M BaCrO4 will begin precipitation.
Among the two, PbCrO4 has a lower Ksp value, that means it will precipitate first from the solution (BaCrO4 remains soluble at this stage).
[CrO4^2-] concentration for BaCrO4 precipitation = Ksp/[Ba2+]
= 2.1 x 10^-10/0.25 = 8.4 x 10^-10 M
11. The solubility of solids that increase in acidic medium are,
C. I and II
These solids in solution forms,
I. Ni(OH)2(s) <==> Ni2+(aq) + 2OH-(aq)
Now, OH- reacts with added acid and forms H2O(l), thereby decreasing concentration of OH- on the right handside. According to LeChatellier's principle, more solid Ni(OH)2 would dissolve to increase OH- concentration and reattain equilibrium state.
II. Mg3(PO4)2(s) <==> 3Mg2+(aq) + 2PO4^3-(aq)
Now, PO4^3- reacts with added acid and forms HPO4^2-(aq), thereby decreasing concentration of PO4^3- on the right handside. According to LeChatellier's principle, more solid Mg3(PO4)2 would dissolve to increase PO4^3- concentration and reattain equilibrium state.
14. For the given mixing of solutions, the true statement would be,
D. Qsp is greater than Ksp for CoCO3 so CoCO3 solid forms
Qsp = [Co2+][CO3^2-]
= (4 x 10^-3 M x 10 ml/50 ml)(0.001 M x 40 ml/50 ml) = 6.4 x 10^-7
So, Qsp > Ksp, therefore, solution is saturated and precipitate would form.
Explanation / Answer
12. If a solution is 0.25 Min Ba2+ and 0.50 MPb+ upon addition of Na2CrO4(s) which solid will precipitate first and at what concentration of Cr042-will the second solid begin to form? Ksp(BaCrO.) 2.1 × 10-10 Ksp(PbCrO.) 2.3x 10-13 A. BaCrO4 will precipitate first, a [Cro,--] = 8.4 × 10-10 M will cause PbCrO4(s) to form. B. BaCrO4 will precipitate first, a [Cr042-] = 4.6 x 10-13 M will cause PbCrO4(s) to form. C. PbCrO4 will precipitate first, a [CrO41 8.4 x 1010 M will cause BaCr04(s) to form. D. PbCrO2 will precipitate first, a [Cr042-] = 4.6 10-13 M will cause BaCrO4(s) to form.
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