Question

Solubility product constant Ksp is also important to predict whether the precipitation will occur under the known condition of ion concentration. This can be done by comparing the solubility quotient (Q) with the value of Ksp of the tested compound. 12. A solution of 0.00016 M lead (II) nitrate, or Pb(NO3)2, was poured into 450 mL of 000023 M sodium sulfate, Na2SO4. Would a precipitate of lead(II)sulfate, PbSO4, be expected to form if 250 mL of the lead nitrate solution were added? Write the equilibrium equation of PbSO4 dissociation in water.

Answer 1

ANSWER:

Data:

• Pb(NO₃)₂ solution
  • 0.00016 M
  • 250 mL = 0.25 L
• Na₂SO₄ solution
  • 0.00023 M
  • 450 mL = 0.45 L
• Total volume of solution (V_T) => 250 mL +450 mL = 700 mL =0.700 L
• K_sp PbSO₄ = 1.6x10^-8

The dissociation of PbSO₄ in water has the following equation

$PbSO_{4}\, (s)\rightleftharpoons Pb^{+2}\, (aq)+SO_{4}^{-2}\, (aq)$

the Ksp for this reaction is:

$K_{sp}=[Pb^{+2}]_{eq}[SO_{4}^{-2}]_{eq}=1.6x10^{-8}$

and the solubility quotient is:

$Q=[Pb^{+2}]_{initial}[SO_{4}^{-2}]_{initial}$

To determine if the PbSO₄ will precipitate when both solutions (Pb(NO₃)₂ and Na₂SO₄) are mixed, we need to find the concentration of Pb⁺² and SO₄^-2:

• All Pb⁺² comes from Pb(NO₃)₂ solution:

$Pb(NO_{3})_{2}\, (aq) \rightarrow Pb^{+2}\, (aq)+2\, NO_{3}^{-}\, (aq)$

according to this, 1 mol of Pb(NO₃)₂ produces 1 mol of Pb⁺², then

$moles\, Pb(NO_{3})_{2}=[Pb(NO_{3})_{2}]\times V_{Pb(NO_{3})_{2}}=moles\, Pb^{+2}$

$moles\, Pb^{+2}=0.00016\, \frac{mol}{L}\times 0.25\, L=\mathbf{4x10^{-5}\, mol}$

the concentration of Pb⁺² when both solutions are mixed is:

$[Pb^{+2}]=\frac{moles\, Pb^{+2}}{V_{T}}=\frac{4x10^{-5}\, mol}{0.700\, L}=\mathbf{5.71x10^{-5}\, M}$

• All SO₄^-2 comes from Na₂SO₄ solution:

$Na_{2}SO_{4}\, (aq) \rightarrow 2\, Na^{+}\, (aq)+ SO_{4}^{-2}\, (aq)$

according to this, 1 mol of Na₂SO₄ produces 1 mol of SO₄^-2, then

$moles\, Na_{2}_SO_{4}=[Na_{2}_SO_{4}]\times V_{Na_{2}_SO_{4}}=moles\, SO_{4}^{-2}$

$moles\, SO_{4}^{-2}=0.00023\, \frac{mol}{L}\times 0.45\, L=\mathbf{1.035x10^{-4}\, mol}$

the concentration of SO₄^-2 when both solutions are mixed is:

$[SO_{4}^{-2}]=\frac{moles\, SO_{4}^{-2}}{V_{T}}=\frac{1.035x10^{-4}\, mol}{0.700\, L}=\mathbf{1.48x10^{-4}\, M}$

Now, we can determine the solubility quotient:

$Q=[Pb^{+2}]_{initial}[SO_{4}^{-2}]_{initial}=5.71x10^{-5}\times 1.48x10^{-4}$

$Q=\mathbf{8.45x10^{-9}}$

In this case, the solubility quotient (Q = 8.45x10^-9) is smaller than Ksp (1.6x10^-8). As Q is smaller than Ksp, the PbSO4 mill not precipitate when the solutions of Na₂SO₄ and Pb(NO₃)₂ are mixed