/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 63 A solution is \(1 \times 10^{-4}... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 63

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 \(K_{\mathrm{sp}}\left[\mathrm{Pb}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]=1 \times 10^{-54}.\)

Short Answer

Expert verified
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 by step solution

01

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.
02

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
03

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
04

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Two different compounds have about the same molar solubility. Do they also have about the same \(K_{\text {sp}}\) value?

Approximately 0.14 g nickel(II) hydroxide, Ni(OH) \(_{2}(s),\) dissolves per liter of water at \(20^{\circ} \mathrm{C}\) . Calculate \(K_{\text { sp }}\) for \(\mathrm{Ni}(\mathrm{OH})_{2}(s)\) at this temperature.

What happens to the \(K_{\mathrm{sp}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.

If \(10.0 \mathrm{mL}\) of \(2.0 \times 10^{-3} M \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}\) is added to 10.0 \(\mathrm{mL}\) of a \(\mathrm{pH}=10.0 \mathrm{NaOH}\) solution, will a precipitate form?

Calculate the solubility of solid \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\left(K_{\mathrm{sp}}=1.3 \times 10^{-32}\right)\) in a \(0.20-M \mathrm{Na}_{3} \mathrm{PO}_{4}\) solution.
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