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Chapter 10: Problem 83

What is the solubility in water for each of the following salts in grams per \(100 \mathrm{~mL}\) ? (a) \(\mathrm{Cu}_{2} \mathrm{~S}\left(K_{\mathrm{sp}}=2.5 \times 10^{-48}\right)\) (b) \(\operatorname{CuS}\left(K_{\mathrm{sp}}=6.3 \times 10^{-36}\right)\)

Short Answer

Expert verified
1.36 x 10鈦宦孤� g/100 mL for Cu鈧係; 2.40 x 10鈦宦光伌 g/100 mL for CuS.

Step by step solution

01

Understand Ksp

The solubility product constant, \(K_{sp}\), tells us about the solubility of a compound in water. It is the product of the concentrations of the ions of the salt, each raised to the power of their coefficients in the balanced equation.
02

Write Dissolution Equation for \(\text{Cu}_2\text{S}\)

\(\text{Cu}_2\text{S}\) dissolves into water according to the equation: \[ \text{Cu}_2\text{S} (s) \rightleftharpoons 2\text{Cu}^{+} (aq) + \text{S}^{2-} (aq) \]Let the solubility be \(s\) moles/L. Then, the equilibrium concentrations are \(2s\) for \(\text{Cu}^{+}\) and \(s\) for \(\text{S}^{2-}\).
03

Set Up Expression for \(K_{sp}\) of \(\text{Cu}_2\text{S}\)

The \(K_{sp}\) expression for \(\text{Cu}_2\text{S}\) is:\[ K_{sp} = [\text{Cu}^+]^2 [\text{S}^{2-}] \]Substitute the concentrations: \[ 2.5 \times 10^{-48} = (2s)^2 \cdot s \] This simplifies to:\[ 2.5 \times 10^{-48} = 4s^3 \]
04

Solve for \(s\) for \(\text{Cu}_2\text{S}\)

Divide by 4:\[ s^3 = \frac{2.5 \times 10^{-48}}{4} \]\[ s^3 = 0.625 \times 10^{-48} \]Solve for \(s\):\[ s = (0.625 \times 10^{-48})^{\frac{1}{3}} \approx 8.54 \times 10^{-17} \text{ mol/L} \]Convert to grams:\(8.54 \times 10^{-17} \text{ mol/L} \times 159.16 \text{ g/mol} = 1.36 \times 10^{-14} \text{ g/L} = 1.36 \times 10^{-12} \text{ g/100 mL}\)
05

Write Dissolution Equation for \(\text{CuS}\)

\(\text{CuS}\) dissolves into water according to the equation: \[ \text{CuS} (s) \rightleftharpoons \text{Cu}^{2+} (aq) + \text{S}^{2-} (aq) \] Let the solubility be \(s\) moles/L. Then, the equilibrium concentrations are \(s\) for both \(\text{Cu}^{2+}\) and \(\text{S}^{2-}\).
06

Set Up Expression for \(K_{sp}\) of \(\text{CuS}\)

The \(K_{sp}\) expression for \(\text{CuS}\) is:\[ K_{sp} = [\text{Cu}^{2+}][\text{S}^{2-}] \]Substitute the concentrations:\[ 6.3 \times 10^{-36} = s^2 \] Solve for \(s\):\[ s = \sqrt{6.3 \times 10^{-36}} \]
07

Solve for \(s\) for \(\text{CuS}\)

Solve for \(s\):\[ s \approx 2.51 \times 10^{-18} \text{ mol/L} \]Convert to grams:\(2.51 \times 10^{-18} \text{ mol/L} \times 95.61 \text{ g/mol} = 2.40 \times 10^{-16} \text{ g/L} = 2.40 \times 10^{-14} \text{ g/100 mL}\)

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Key Concepts

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

Dissolution Equation
The dissolution equation is key to understanding how ionic compounds dissolve in water. For salts like CuS and Cu鈧係, this equation helps determine how their components break apart in water.
For instance, when Cu鈧係 dissolves, it separates into its ionic parts: 2 Cu鈦� and one S虏鈦�. This decomposition is represented by the dissolution equation:
  • Cu鈧係 (s) 鈬� 2 Cu鈦�(aq) + S虏鈦�(aq)
Similarly, CuS dissolves into Cu虏鈦� and S虏鈦� ions, represented as:
  • CuS (s) 鈬� Cu虏鈦�(aq) + S虏鈦�(aq)
These equations provide the basis for calculating solubility and setting up the solubility product expression (Ksp). They show how many moles of ions are produced per mole of salt dissolved.
Solubility Calculation
Solubility refers to the maximum amount of solute that can dissolve in a solvent at equilibrium. Calculating solubility involves working with the solubility product constant ( K_{sp} ).
Let's break this calculation down:
  • First, write the dissolution equation to understand the ions produced.
  • Next, set up the K_{sp} expression using the concentrations of these ions at equilibrium.
  • Substitute the equilibrium concentrations into the K_{sp} expression and solve for s (solubility in mol/L).
  • Finally, convert s into grams per given volume using the molar mass of the compound.
For example, with Cu鈧係, the solubility calculation starts from its dissolution equation. Knowing the K_{sp} value, you can derive s through a series of substitutions and simplifications, leading to a molar solubility which can then be converted into grams per 100 mL.
Equilibrium Concentration
Equilibrium concentration is fundamental in solubility calculations. It represents the concentration of ions in a saturated solution when equilibrium is reached.
In a dissolution process:
  • Equilibrium is the state where the rate of dissolution equals the rate of precipitation.
  • The concentrations of ions at this point are used in the K_{sp} expression.
  • For example, the equilibrium concentrations of ions for Cu鈧係 are 2s for Cu鈦� and s for S虏鈦�.
  • Using these, the K_{sp} expression becomes (2s)^2 imes s .
This means, once you substitute the equilibrium values into the K_{sp} expression, you can solve for s . Understanding equilibrium concentrations is crucial to finding how much of a salt can dissolve in a solution.

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Most popular questions from this chapter

Suppose that the reaction quotient \(\left(Q_{\mathrm{c}}\right)\) for the following reaction at some moment in time is \(1.0 \times 10^{-8}\) and the equilibrium constant for the reaction \(\left(K_{\mathrm{c}}\right)\) at the same temperature is \(3 \times 10^{-7}\) $$ 2 \mathrm{NO}_{2}(g) \rightleftarrows 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ Which of the following is a valid conclusion? (a) The reaction is at equilibrium. (b) The reaction must shift toward the products to reach equilibrium. (c) The reaction must shift toward the reactants to reach equilibrium.

Explain how the rates of the forward and reverse reactions change as the reaction between \(\mathrm{ClNO}_{2}\) and NO proceeds to equilibrium. Assume that no \(\mathrm{ClNO}\) or \(\mathrm{NO}_{2}\) are present initially.

Without detailed equilibrium calculations, estimate the concentrations of \(\mathrm{NO}\) and \(\mathrm{NOCl}\) at equilibrium if a mixture that was initially \(0.50 \mathrm{M}\) in \(\mathrm{NO}\) and \(0.10 \mathrm{M}\) in \(\mathrm{Cl}_{2}\) combined to form nitrosyl chloride, NOCl. $$ \begin{array}{c} 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftarrows 2 \mathrm{NOCl}(g) \\ K_{\mathrm{c}}=2.1 \times 10^{3}(\text { at } 500 \mathrm{~K}) \end{array} $$

What is the solubility of silver sulfide in water in grams per \(100 \mathrm{~mL}\) if the solubility product for \(\mathrm{Ag}_{2} \mathrm{~S}\) is \(6.3 \times 10^{-50} ?\)

Without detailed equilibrium calculations, estimate the equilibrium concentration of \(\mathrm{SO}_{3}\) when a mixture of \(0.100 \mathrm{~mol}\) of \(\mathrm{SO}_{2}\) and \(0.050 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) flask at \(300^{\circ} \mathrm{C}\) combine to form \(\mathrm{SO}_{3}\). Assume that \(K_{\mathrm{c}}=\) \(6.3 \times 10^{9}\) for this reaction at \(300^{\circ} \mathrm{C}\) $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g) $$
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