What will be the result when 15.0 mL of 0.040 M Pb^2+ is mixed with 15.0 mL of 0

Question

What will be the result when 15.0 mL of 0.040 M Pb^2+ is mixed with 15.0 mL of 0.040 M Cl? K_Sp = 1.7 times 10^5 A clear solution with no precipitate will result. Solid PbCl_2 will precipitate and excess Pb^2+ ions will remain in solution. Solid PbCl_2 will precipitate and excess Cl^- ions will remain in solution. Solid PbCl_2 will precipitate and no excess ions will remain in solution. Copper(II) hydroxide, Cu(OH)_2(s), has K_sp = 2.2 times 10^-20. For the reaction below, K = 4.0 times 10^-7. What is K_f for Cu(NH_3)_4^2+? 8.8 times 10^-27 5.5 times 10^-14 1.8 times 10^13 1.1 times 10^26 The following anions can be separated by precipitation as silver salts: Cl^-, Br', I^-, F. If Ag^+ is added to a solution containing the four anions, each at a concentration of 0.10 M, in what order will they precipitate?

Explanation / Answer

Ans. 19

The solubility product Ksp, for PbCl2 is 1.7 × 10–5.

PbCl2(s) <==> Pb2+(aq) + 2Cl-(aq)

Ksp = [Pb2+][Cl-]^2 = (x) (2x) 2
So the solubility of PbCl2 is 0.016 M

In the given problem 15 ml of 0.040 M Pb2+ is mixed with 15 ml of 0.040 M Cl- so a clear solution with no precipitate will result.

Ans. 20

Ksp of Cu(OH)2 (s) = 2.2 x 10-20as given

Cu(OH)2 (s) + NH3 aq ==>      Cu(NH3)42+aq    + OH-aq for this reaction K= 4 x 10 -7.

K net= Ksp x Kf

Kf= K net/ Ksp = 4 x 10 -7 /2.2 x 10-20 = 1.8 x 10 13

Option ‘C’ is correct

Ans. 21

In the solution the precipitation will take place in following order on the basis of their solubility product.

AgI==> AgBr==>AgCl==>AgF

Option b is correct

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