Suppose 50.00 mL of a 1 × 10^-7 M solution of lead(II)nitrate is mixed with 50.00 mL of a 1 × 10^-8 solution of sodium phosphate.Which of the following statements is true? For lead(II)phosphate,K_sp = 1 × 10^-44. A)A precipitate forms because Q_c K_sp. C)A precipitate forms because Q_c K_sp. D)No precipitate forms because Q_c = K_sp. E)No precipitate forms because Q_c

Question

Suppose 50.00 mL of a 1 × 10^-7 M solution of lead(II)nitrate is mixed with 50.00 mL of a 1 × 10^-8 solution of sodium phosphate.Which of the following statements is true? For lead(II)phosphate,K_sp = 1 × 10^-44. A)A precipitate forms because Q_c < K_sp. B)No precipitate forms because Q_c > K_sp. C)A precipitate forms because Q_c > K_sp. D)No precipitate forms because Q_c = K_sp. E)No precipitate forms because Q_c < K_sp.

Quizplus · Accepted Answer

The reaction between lead(II) nitrate and sodium phosphate can be represented as:

Pb(NO3)2(aq) + Na3PO4(aq) → Pb3(PO4)2(s) + 3NaNO3(aq)

The balanced equation shows that 1 mole of Pb(NO3)2 reacts with 1 mole of Na3PO4 to produce 1 mole of Pb3(PO4)2.

To determine if a precipitate will form, we need to compare the ion product, Qsp, of Pb3(PO4)2 with its solubility product constant, Ksp.

Qsp = [Pb2+][PO43-]^2

We can calculate the concentrations of Pb2+ and PO43- in the solution after mixing the two solutions.

[Pb2+] = (1x10^-7 M) x (50.00 mL / 100.00 mL) = 5.0 x 10^-8 M

[PO43-] = (1x10^-8 M) x (50.00 mL / 100.00 mL) = 5.0 x 10^-9 M

Substituting these values into the expression for Qsp gives:

Qsp = (5.0 x 10^-8)(5.0 x 10^-9)^2 = 1.25 x 10^-25

Comparing Qsp to the given solubility product constant, Ksp, we see that:

Qsp > Ksp

This means that the ion product is greater than the solubility product, indicating that the reaction quotient is not at equilibrium and that a precipitate will form until Qsp = Ksp. Therefore, choice C is the best answer.