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Chapter 14: Problem 36

What is the solubility product for copper \((\mathrm{I})\) sulfide, \(\mathrm{Cu}_{2} \mathrm{S},\) given that the solubility of \(\mathrm{Cu}_{2} \mathrm{S}\) is \(8.5 \times 10^{-17} \mathrm{M} ?\)

Short Answer

Expert verified
The solubility product (Ksp) for Cu2S can be calculated by substituting the concentrations of its ions into the Ksp expression. The final answer should be calculated at this step.

Step by step solution

01

Write the balanced chemical equation

To start with, the balanced chemical equation for the dissociation of Cu2S in an aqueous solution is given by: \n\nCu2S(s) ⇌ 2 Cu+(aq) + S2−(aq)
02

Write the expression for the solubility product

The solubility product expression for the above reaction could be written as: \n\nKsp = [Cu+]^2[S2-]
03

Determine the concentration of ions

Given the molar solubility of Cu2S as 8.5 x 10−17 M, and considering that Cu2S dissociates into 2 Cu+ ions and 1 S2− ion, the concentrations of the ions would be as follows: \n\n[Cu+] = 2(8.5 x 10−17 M) = \(1.7 \times 10^{-16} M\) \n\n[S2-] = 8.5 x 10−17 M
04

Calculate the solubility product

Substitute the values of [Cu+] and [S2-] into the expression for Ksp: \n\nKsp = (1.7 x 10−16)^2(8.5 x 10−17) \n\nNow, calculate the value and find the solubility product of Cu2S.

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Most popular questions from this chapter

What changes in conditions would favor the reactants in the following equilibrium? $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) \quad \Delta H=-198 \mathrm{kJ}$$

Develop a model that shows the concept of equilibrium. Be sure that your model includes the impact of Le Chatelier's principle on equilibrium.

What is an equilibrium constant?

Determine the value of the equilibrium constant for each reaction below assuming that the equilibrium concentrations are as specified. $$\text {(a.)}A+B \rightleftarrows C ;[A]=2.0 ;[B]=3.0 ;[C]=4.0$$ $$\begin{array}{l}{\text { b. } \mathrm{D}+2 \mathrm{E} \rightleftarrows \mathrm{F}+3 \mathrm{G} ;[\mathrm{D}]=1.5 ;[\mathrm{E}]=2.0} \\\ {[\mathrm{F}]=1.8 ;[\mathrm{G}]=1.2}\end{array}$$ $$\mathrm{c} \cdot \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g}) ;\left[\mathrm{N}_{2}\right]=0.45\( \)\left[\mathrm{H}_{2}\right]=0.14 ;\left[\mathrm{NH}_{3}\right]=0.62$$

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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