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Chapter 17: Problem 50

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\), is the same, \(4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} .(\mathbf{a})\) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?(\mathbf{c})\) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2}\), what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

Short Answer

Expert verified
(a) Salt MZ鈧� has the larger solubility product constant, with Ksp鈧� = \(128 \times 10^{-12}\) while Ksp鈧� = \(16 \times 10^{-8}\) for salt MA. (b) Neither salt has a higher concentration of M虏鈦� ions in the saturated solution as both their solubilities are the same. (c) The equilibrium concentration of M虏鈦� ions after mixing equal volumes of saturated solutions of MA and MZ鈧� is \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\).

Step by step solution

01

(a) Determine the solubility product constant

The solubility product constant (Ksp) is calculated using the solubility equilibrium expressions for the given salts. For salt MA: \(MA \rightleftharpoons M^{2+} + A^-\) Ksp鈧� = [M虏鈦篯[A鈦籡 For salt MZ鈧�: \(MZ_2 \rightleftharpoons M^{2+} + 2Z^-\) Ksp鈧� = [M虏鈦篯[Z鈦籡虏 Given the solubility, s = \(4 \times 10^{-4}\, \mathrm{mol\, L^{-1}} \) For salt MA: [M虏鈦篯 = [A鈦籡 = s = \(4 \times 10^{-4} \mathrm{mol\, L^{-1}}\) Ksp鈧� = (\(4 \times 10^{-4}\))^2 = \(16 \times 10^{-8}\) For salt MZ鈧�: [M虏鈦篯 = s = \(4 \times 10^{-4} \mathrm{mol\, L^{-1}}\) [Z鈦籡 = 2s = \(8 \times 10^{-4} \mathrm{mol\, L^{-1}}\) Ksp鈧� = (\(4 \times 10^{-4}\))(\(8 \times 10^{-4}\))^2 = \(128 \times 10^{-12}\) Comparing the solubility product constants, Ksp鈧� > Ksp鈧�. Therefore, salt MZ鈧� has the larger numerical value for the solubility product constant.
02

(b) Determine the higher concentration of M虏鈦� in the saturated solution

The solubility of both salts is given as the same, so the concentration of M虏鈦� ions in the saturated solution of both salts would also be the same. Thus, neither of them has a higher concentration of M虏鈦� ions in the saturated solution.
03

(c) Determine the equilibrium concentration of M虏鈦� after mixing the solutions

When mixing equal volumes of saturated solutions of MA and MZ鈧�, the total volume doubles, and so does the number of moles of each ion. However, the concentration of each ion gets halved. For the mixed solution: [M虏鈦�'] = \(\frac{1}{2}\) [M虏鈦篯 = \(\frac{1}{2}\) (\(4 \times 10^{-4}\, \mathrm{mol\, L^{-1}}\)) = \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\) Therefore, the equilibrium concentration of M虏鈦� ions after mixing the solutions is \(2 \times 10^{-4} \mathrm{mol\, L^{-1}}\).

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Key Concepts

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

Saturated Solution
A saturated solution is defined as a solution in which the maximum amount of solute is dissolved at a given temperature. In other words, no more solute can dissolve in the solvent without changing the temperature or solvent amount. In the case of the two slightly soluble salts, MA and MZ鈧�, a saturated solution occurs when the amount of each salt that can dissolve has been reached.
When a solution reaches saturation, the rate at which the salt dissolves is equal to the rate at which it precipitates. This establishes a dynamic equilibrium. For MA and MZ鈧�, each saturates the solution at a solubility of \(4 \times 10^{-4} \ \text{mol/L}\). Beyond this point, any additional salt will remain undissolved.
Solubility Equilibrium
Solubility equilibrium refers to the state where the process of dissolving and the precipitation of a solute occur at equal rates, maintaining the concentration of dissolved ions fixed in a saturated solution. It is represented by equilibrium expressions that vary depending on the dissociation form of the salt.
For example, with the salts MA and MZ鈧�:
  • MA dissociates into \(\text{M}^{2+}\) and \(\text{A}^-\).
  • MZ鈧� dissociates into \(\text{M}^{2+}\) and \(2\text{Z}^-\).
The solubility product constant \(K_{sp}\) quantifies the equilibrium state for slightly soluble salts. It is calculated as:
MA: \[ K_{sp1} = [\text{M}^{2+}] [\text{A}^-] \]
MZ鈧�: \[ K_{sp2} = [\text{M}^{2+}] [\text{Z}^-]^2 \]
These expressions help us understand the concentration dynamics of the ions at equilibrium.
Ionic Concentration
Ionic concentration in a solution describes how many ions are present in a given volume of the solution. In a saturated solution, the ionic concentration rises until solubility equilibrium is reached.
Despite MA and MZ鈧� having the same solubility of \(4 \times 10^{-4} \ \text{mol/L}\), their dissociation affects ionic concentrations differently. For MA, the concentration of \(\text{M}^{2+}\) is equal to its solubility. However, in MZ鈧�, the dissociation doubles the concentration of \(\text{Z}^-\) ions.
When the solutions of MA and MZ鈧� are mixed, the resulting solution sees a balance due to dilution, halving the ionic concentration. For \(\text{M}^{2+}\), this means its concentration reduces to \(2 \times 10^{-4} \ \text{mol/L}\), assuming equal volumes of the solutions are mixed. Understanding ionic concentration is key to predicting how substances will behave in saturated solutions.

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Most popular questions from this chapter

Write the expression for the solubility-product constant for each of the following ionic compounds: \(\mathrm{BaCrO}_{4}, \mathrm{CuS}, \mathrm{PbCl}_{2}\) and \(\mathrm{LaF}_{3} .\)

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16}\). The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}{ }^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

Consider the equilibrium $$ \mathrm{B}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HB}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ Suppose that a salt of \(\mathrm{HB}^{+}(a q)\) is added to a solution of \(\mathrm{B}(a q)\) at equilibrium. (a) Will the equilibrium constant for the reaction increase, decrease, or stay the same? (b) Will the concentration of \(\mathrm{B}(a q)\) increase, decrease, or stay the same? (c) Will the pH of the solution increase, decrease, or stay the same?

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\) ) reacts with acids in foods to form carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(g)\) forms bubbles and causes the cake to rise. \((\mathbf{a})\) A rule of thumb in baking is that \(1 / 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\). Write the chemical equation for this neutralization reaction. (b) The density of baking soda is \(2.16 \mathrm{~g} / \mathrm{cm}^{3}\). Calculate the concentration of lactic acid in one cup of sour milk (assuming the rule of thumb applies), in units of mol/L. (One cup \(=236.6 \mathrm{~mL}=48\) teaspoons \() .(\mathbf{c})\) If \(1 / 2\) teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at a pressure of \(101.3 \mathrm{kPa}\), in an oven set to \(177^{\circ} \mathrm{C}\).
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