Problem 63 - Chemistry

Question

A solution is \(1 \times 10^{-4} M\) in \(\mathrm{NaF}\), \(\mathrm{Na}_{2} \mathrm{S},\) and \(\mathrm{Na}_{3} \mathrm{PO}_{4} .\) What would be the order of precipitation as a source of \(\mathrm{Pb}^{2+}\) is added gradually to the solution? The relevant \(K_{\mathrm{sp}}\) values are \(K_{\mathrm{sp}}\left(\mathrm{PbF}_{2}\right)=4 \times 10^{-8}, K_{\mathrm{sp}}(\mathrm{PbS})=7 \times 10^{-29},\) and \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}.\)

Step-by-Step Solution

Answer

The order of precipitation when a source of Pb^2+ is added gradually to the solution containing NaF, Na2S, and Na3PO4 is: 1. Pb3(PO4)2 2. PbF2 3. PbS

Step 1: Calculate the concentration of anions in the solution

Given the concentration of NaF, Na2S, and Na3PO4 is 1 x 10^-4 M, the concentration of F^-, S^2-, and PO4^3- in the solution will also be 1 x 10^-4 M.

Step 2: Calculate the saturation quotient (Q) for PbF2, PbS, and Pb3(PO4)2

Use the formula Q = [Pb^2+][A^n-]: For PbF2: Q = [Pb^2+][F^-] = [Pb^2+](1 x 10^-4 M) For PbS: Q = [Pb^2+][S^2-] = [Pb^2+](1 x 10^-4 M) For Pb3(PO4)2: Q = [Pb^2+]^3[PO4^3-]^2 = ([Pb^2+]^3)(1 x 10^-4 M)^2

Step 3: Compare calculated Q values to the given Ksp values

The compound with the smallest Q value above its Ksp will precipitate first. We have: Ksp(PbF2) = 4 x 10^-8 = [Pb^2+](1 x 10^-4 M) Ksp(PbS) = 7 x 10^-29 = [Pb^2+](1 x 10^-4 M) Ksp(Pb3(PO4)2) = 1 x 10^-54 = ([Pb^2+]^3)(1 x 10^-4 M)^2 Calculate the [Pb^2+] values for which the Q values are equal to the Ksp values: [Pb^2+]_PbF2 = 4 x 10^-4 M [Pb^2+]_PbS = 7 x 10^-25 M [Pb^2+]_Pb3(PO4)2 = 1 x 10^-16 M

Step 4: Determine the order of precipitation

Since the [Pb^2+] required to reach the Ksp is smallest for Pb3(PO4)2, it will precipitate first. The next smallest [Pb^2+] value corresponds to PbF2, and the largest to PbS. So, the order of precipitation is as follows: 1. Pb3(PO4)2 2. PbF2 3. PbS

Key Concepts

Ksp (Solubility Product Constant)

The solubility product constant, or \( K_{sp} \), is essential for understanding precipitation reactions. It represents the maximum amount of a solid that can dissolve in a solution to form a saturated solution under equilibrium.
The \( K_{sp} \) is specific to each compound and indicates how soluble that compound is. For example, in the given exercise:

• \( K_{sp} \) for \( \text{PbF}_2 \) is \(4 \times 10^{-8} \)
• \( K_{sp} \) for \( \text{PbS} \) is \(7 \times 10^{-29} \)
• \( K_{sp} \) for \( \text{Pb}_3(\text{PO}_4)_2 \) is \(1 \times 10^{-54} \)

A smaller \( K_{sp} \) value means the compound is less soluble; thus, it will precipitate out of the solution first if more cations or anions are added. Recognizing \( K_{sp} \) values helps predict which compounds will form a precipitate first.

Anion Concentration

Anions in a solution interact with cations to form various compounds. Their concentration is crucial in determining the likelihood of a precipitation reaction.
In this exercise, the concentrations of \( \text{F}^- \), \( \text{S}^{2-} \), and \( \text{PO}_4^{3-} \) are each \(1 \times 10^{-4} \text{ M} \), as they are derived from \( \text{NaF} \), \( \text{Na}_2\text{S} \), and \( \text{Na}_3\text{PO}_4 \) respectively.

• The higher the anion concentration, the more immediate the possibility of an ion reaching its \( K_{sp} \)
• These concentrations are crucial in calculating the saturation quotient \( Q \)

By understanding the role of anion concentration, you can better predict which precipitates will form when a common ion is added.

Saturation Quotient (Q)

The saturation quotient \( Q \) helps determine if a solution is saturated, unsaturated, or supersaturated.
It is calculated using the concentrations of ions in the solution:

• For \( \text{PbF}_2 \), \( Q = [\text{Pb}^{2+}][\text{F}^-]^2 \)
• For \( \text{PbS} \), \( Q = [\text{Pb}^{2+}][\text{S}^{2-}] \)
• For \( \text{Pb}_3 (\text{PO}_4 )_2 \), \( Q = [\text{Pb}^{2+}]^3[\text{PO}_4^{3-}]^2 \)

Comparing \( Q \) with \( K_{sp} \) values tells us if precipitation will occur:

• If \( Q < K_{sp} \), the solution is unsaturated (no precipitation)
• If \( Q = K_{sp} \), the solution is saturated (in equilibrium)
• If \( Q > K_{sp} \), the solution is supersaturated (precipitation occurs)

Understanding \( Q \) allows you to predict when and how precipitates will form as the solution conditions change.