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Chapter 17: Problem 78

The solubility product constant for lead(II) arsenate \(\left(\mathrm{Pb}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\right)\) is \(4.0 \times 10^{-36}\) at 298 \(\mathrm{K}\) . Calculate the molar solubility of the compound at this temperature.

Short Answer

Expert verified
The molar solubility of lead(II) arsenate at 298 K is approximately \(7.3 \times 10^{-9}\) mol/L.

Step by step solution

01

Write the balanced equation for the dissolution of lead(II) arsenate

The balanced equation for the dissolution of lead(II) arsenate is given as follows: \(Pb_3(AsO_4)_2 \rightleftharpoons 3Pb^{2+} + 2AsO_4^{3-}\)
02

Express the concentration terms in terms of the molar solubility

Let's denote molar solubility as "s". Based on the stoichiometry of the balanced equation above, we know that: - For every molecule of Pb鈧�(AsO鈧�)鈧� dissolved, 3 molecules of Pb虏鈦� ions are formed. - For every molecule of Pb鈧�(AsO鈧�)鈧� dissolved, 2 molecules of AsO鈧劼斥伝 ions are formed. So, the concentration of Pb虏鈦� ions = 3s, and the concentration of AsO鈧劼斥伝 ions = 2s.
03

Write the expression for Ksp using the concentration terms

The Ksp expression for the balanced equation can be written as: \(K_{sp} = [Pb^{2+}]^3[AsO_4^{3-}]^2\) Substitute the concentration terms with the expressions in terms of solubility (s): \(K_{sp} = (3s)^3(2s)^2\)
04

Solve for the molar solubility in mol/L

Given the value of Ksp for lead(II) arsenate to be 4.0 脳 10鈦宦斥伓, we substitute this value into the Ksp expression above and solve for "s": \(4.0 \times 10^{-36} = (3s)^3(2s)^2\) Now let's solve for "s": \(4.0 \times 10^{-36} = 108s^5\) Divide both sides by 108: \(s^5 = \frac{4.0 \times 10^{-36}}{108}\) Then, find the fifth root of both sides: \(s = \sqrt[5]{\frac{4.0 \times 10^{-36}}{108}}\) \(s \approx 7.3 \times 10^{-9}\) The molar solubility of lead(II) arsenate at 298 K is approximately \(7.3 \times 10^{-9}\) mol/L.

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Key Concepts

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

Solubility Product Constant
The solubility product constant, often abbreviated as "Ksp," is a special type of equilibrium constant. It is specific to the solubility of ionic compounds in a solution. When an ionic compound dissolves in water, it disassociates into its constituent ions.
The Ksp is the product of the concentrations of these ions each raised to the power of their stoichiometric coefficients in the balanced equation. For instance, if the equation for a compound dissolving is \( A_xB_y\rightleftharpoons xA^{n+} + yB^{m-} \), the expression for Ksp is:
  • \( K_{sp} = [A^{n+}]^x[B^{m-}]^y \).
A low Ksp value generally means the compound is not very soluble, while a high Ksp value indicates higher solubility. The Ksp provides insight into whether a precipitate will form in a solution.
Lead(II) Arsenate
Lead(II) arsenate is a chemical compound with the formula \( \text{Pb}_3(\text{AsO}_4)_2 \). It consists of lead and arsenate ions.
In the context of solubility and Ksp calculations, lead(II) arsenate undergoes dissolution, breaking down into lead ions \( \text{Pb}^{2+} \) and arsenate ions \( \text{AsO}_4^{3-} \).
This compound is known for its low solubility in water, as illustrated by its very small Ksp value of \( 4.0 \times 10^{-36} \).
  • This indicates that only a tiny amount of lead(II) arsenate can dissolve in water before reaching equilibrium.
  • The low solubility is significant in environmental and health contexts, as lead compounds can have toxic effects.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In the study of solubility, stoichiometry helps us predict the concentrations of ions based on the dissolution of a compound.
When lead(II) arsenate dissolves, stoichiometry tells us that one molecule of \( \text{Pb}_3(\text{AsO}_4)_2 \) produces three \( \text{Pb}^{2+} \) ions and two \( \text{AsO}_4^{3-} \) ions.
  • This relationship can be represented as: for every mole of \( \text{Pb}_3(\text{AsO}_4)_2 \) that dissolves, three moles of lead ions and two moles of arsenate ions are formed.
  • This is crucial as it allows us to write out the expression for the concentrations of ions: \( [\text{Pb}^{2+}] = 3s \) and \( [\text{AsO}_4^{3-}] = 2s \), where \( s \) is the molar solubility.
Ksp Expression
The Ksp expression is a critical part of determining how soluble an ionic compound is in a solution. For the dissolution of lead(II) arsenate, the Ksp expression takes into account both the balanced chemical equation and the stoichiometry of the ions produced.
From the balanced equation \( \text{Pb}_3(\text{AsO}_4)_2 \rightleftharpoons 3\text{Pb}^{2+} + 2\text{AsO}_4^{3-} \), we construct the Ksp expression as follows:
  • \( K_{sp} = [\text{Pb}^{2+}]^3 [\text{AsO}_4^{3-}]^2 \)
By substituting \( [\text{Pb}^{2+}] = 3s \) and \( [\text{AsO}_4^{3-}] = 2s \) into the Ksp formula, we obtain:
  • \( K_{sp} = (3s)^3 (2s)^2 \)
This simplifies to \( K_{sp} = 108s^5 \), linking the mathematical solution to the chemical properties of lead(II) arsenate.

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Most popular questions from this chapter

Explain how you would regulate the pressure to favor the products in the following equilibrium system. $$ \mathrm{MgCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$

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