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Chapter 14: Problem 37

The ionic substance \(\mathrm{T}_{3} \mathrm{U}_{2}\) ionizes to form \(\mathrm{T}^{2+}\) and \(\mathrm{U}^{3-}\) ions. The solubility of \(\mathrm{T}_{3} \mathrm{U}_{2}\) is \(3.77 \times 10^{-20} \mathrm{mol} / \mathrm{L}\) . What is the value of the solubility-product constant?

Short Answer

Expert verified
The solubility-product constant for the substance \(\mathrm{T}_{3} \mathrm{U}_{2}\) is \(7.06 \times 10^{-97}\)

Step by step solution

01

Write the dissolution process

When \(\mathrm{T}_{3} \mathrm{U}_{2}\) dissolves in water, it splits into its ions according to the following balanced chemical equation: \(\mathrm{T}_{3} \mathrm{U}_{2} \rightarrow 3\mathrm{T}^{2+} + 2\mathrm{U}^{3-}\)
02

Determine the concentrations of ions

According to the stoichiometry of the reaction, each dissolved formula unit of \(\mathrm{T}_{3} \mathrm{U}_{2}\) results in three \(\mathrm{T}^{2+}\) ions and two \(\mathrm{U}^{3-}\) ions. Hence, the concentration of \(\mathrm{T}^{2+}\) in the solution is \(3 \times 3.77 \times 10^{-20} \mathrm{mol} / \mathrm{L} = 1.131 \times 10^{-19} \mathrm{mol} / \mathrm{L}\) and the concentration of \(\mathrm{U}^{3-}\) is \(2 \times 3.77 \times 10^{-20} \mathrm{mol} / \mathrm{L} = 7.54 \times 10^{-20} \mathrm{mol} / \mathrm{L}\)
03

Use the solubility-product expression to calculate the solubility-product constant

The solubility-product expression is: \(K_{sp} = [\mathrm{T}^{2+}]^3 [\mathrm{U}^{3-}]^2\). Substituting the calculated concentrations into the solubility product expression, we get \(K_{sp} = (1.131 \times 10^{-19})^3 \times (7.54 \times 10^{-20})^2 = 7.06 \times 10^{-97}\)

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Most popular questions from this chapter

A student wrote, "The larger the equilibrium constant, the greater the rate at which reactants convert to products." How was he wrong?

Changes in the concentrations of the reactants and products at equilibrium have no effect on the value of the equilibrium constant. Explain this statement.

When does a pressure change affect a chemical equilibrium?

Imagine the following hypothetical reaction, taking place in a sealed, rigid container, to be neither exothermic nor endothermic. $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftarrows \mathrm{C}(g) \quad \Delta H=0$$ Would an increase in temperature favor the forward reaction or the reverse reaction? (Hint: Recall the gas laws.)

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) can be prepared in the presence of a catalyst by the reaction of \(\mathrm{H}_{2}\) and \(\mathrm{CO}\) at high temperatures according to the following equation: $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}(g)$$ What is the concentration of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in moles per liter if the concentration of \(\mathrm{H}_{2}=0.080 \mathrm{mol} / \mathrm{L}\) , the concentration of \(\mathrm{CO}=0.025 \mathrm{mol} / \mathrm{L},\) and \(K_{e q}=290\) at 700 \(\mathrm{K} ?\)
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