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Chapter 14: Problem 25

The saturation solubility of \(\mathrm{Ag}_{2} \mathrm{~S}\) at \(25^{\circ} \mathrm{C}\) is \(1.14 \times 10^{-17} \mathrm{M}\). What are the equilibrium concentrations of the cation and anion?

Short Answer

Expert verified
The equilibrium concentrations of the cation (Ag⁺) and anion (S²⁻) are \(2.84 \times 10^{-6} M\) and \(1.42 \times 10^{-6} M\), respectively.

Step by step solution

01

Write the dissolution reaction and its equilibrium expression

First, we need to write the balanced chemical equation for the dissolution of Ag2S in water and then write the equilibrium expression (Ksp) for the reaction. The balanced chemical equation for the dissolution of Ag2S is: \(Ag_{2}S_{(s)} \rightleftharpoons 2Ag_{(aq)}^{+} + S_{(aq)}^{2-}\) The equilibrium expression (Ksp) for this reaction can be obtained by taking the product of the concentrations of the ions to the power of their stoichiometric coefficients: \(K_{sp} = [Ag^{+}]^{2} [S^{2-}]\)
02

Relate solubility to the equilibrium concentrations of the ions

We are given that the saturation solubility of Ag₂S is 1.14 x 10⁻¹⁷ M. Let's denote this solubility as 's'. When Ag₂S dissolves in water, it forms 2 moles of Ag⁺ ions and 1 mole of S²⁻ ion. Therefore, we can relate the solubility to the equilibrium concentrations of the ions as follows: \(Ag^{+}: [Ag^{+}] = 2s\) \(S^{2-}: [S^{2-}] = s\)
03

Substitute the equilibrium concentrations into the Ksp expression and solve for 's'

Now, we substitute the equilibrium concentrations that we just found, in terms of 's', into the Ksp expression: \(K_{sp} = (2s)^{2}(s)\) We know that the saturation solubility of Ag₂S is 1.14 x 10⁻¹⁷ M, so we can equate this to 's': \(1.14 \times 10^{-17} = (2s)^{2}(s)\) Now we need to solve this equation for 's' to find the solubility of Ag₂S.
04

Solve the equation for 's'

We can solve this equation by simplifying and finding the value of 's': \(1.14 \times 10^{-17} = 4s^3\) Dividing both sides by 4: \(s^3 = \frac{1.14 \times 10^{-17}}{4}\) Taking the cube root of both sides: \(s = (\frac{1.14 \times 10^{-17}}{4})^{\frac{1}{3}}\) After solving, we find that: \(s = 1.42 \times 10^{-6} M\)
05

Find the equilibrium concentrations of the cation and anion

Now that we have found the value of 's', we can determine the equilibrium concentrations of Ag⁺ and S²⁻: \(Ag^{+}: [Ag^{+}] = 2s = 2(1.42 \times 10^{-6}) = 2.84 \times 10^{-6} M\) \(S^{2-}: [S^{2-}] = s = 1.42 \times 10^{-6} M\) So, the equilibrium concentrations of the cation (Ag⁺) and anion (S²⁻) are 2.84 x 10⁻⁶ M and 1.42 x 10⁻⁶ M, respectively.

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Most popular questions from this chapter

Suppose a reaction has a \(K_{\text {eq }}\) value of \(2.05\). When we write the reaction, can we use a single arrow to the right instead of a double set of equilibrium arrows? Explain your answer.

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Suppose the reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightleftarrows 2 \mathrm{AB}\) proceeds via a one-step mechanism involving a collision between one \(\mathrm{A}_{2}\) molecule and one \(\mathrm{B}_{2}\) molecule. Suppose also that this reaction is reversible, and that the forward reaction is inherently much faster than the reverse reaction. (a) Does the equilibrium lie to the left or to the right? Explain your choice in terms of the reactant and product concentrations necessary to establish equal forward and reverse rates. (b) Does an analysis in terms of the relationship \(K_{\mathrm{eq}}=k_{\mathrm{f}} / k_{\mathrm{r}}\) yield the same answer as in (a)? Explain.

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