Although PbCl_2 has solubility in water at room temperature, the solubility of P

Question

Although PbCl_2 has solubility in water at room temperature, the solubility of PbCl_2 can be increased by raising the temperature of the solution or by increasing the amount off water. PbCI_2(s) Pb^2+ (aq) + 2Cl^- (aq) (a) Write the equilibrium expression for this reaction. (b) If 5.0 ml off 0.30 M Pb(NO_3)_2 solution is added to a clean test tube, how many moles off Pb^2+ are present? (c) If 2.0 ml off 0.30 M HCI is then added to the test tube, white flakes off PbCI_2 precipitates out. The precipitate is dissolved by heating the solution in hot water. Did the K_sp increase or decrease upon heating? (d) The solution was cooled and the precipitate re appeared. Five (5.0) mLs of water was needed to re-dissolve the precipitate. At the end, what is the (Pb^2+]? What is the [CI^+]? What is the value for K_sp?

Explanation / Answer

(a)

Equilibrium constant expression, K = [Pb2+][Cl-]2/[PbCl2]

K*[PbCl2] = [Pb2+][Cl-]2

but cocentration of PbCl2 solid is always constant, so, K*[PbCl2] = Ksp

Ksp = [Pb2+][Cl-]2

(b) Pb(NO3)2 (aq.) -----------> Pb2+ (aq.) + 2 NO3- (aq.)

Moles of Pb(NO3)2 = moles of Pb2+ = 0.30 * 5.0 / 100 = 0.0015 mol

(c)

HCl (aq.) ----------> H+ (aq) + Cl- (aq.)

Moles of HCl = Moles of H+ = 0.30 * 2.0 / 1000 = 0.00060 mol

As the precipitate is dissolved on increasing temperature, the concentration of Pb2+ and Cl- ions will increase and thus Ksp also increases.

(d) Final, [Pb2+] = 0.0015 * 1000 / (5.0+2.0+5.0) = 0.125 M

final [Cl-] = 0.00060 * 1000 / 12 = 0.050 M

therefore,

Ksp = [Pb2+][Cl-]2

Ksp = (0.125)(0.050)2

Ksp = 3.125 * 10-4 M3

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