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Chapter 8: Problem 135

What is the \(\mathrm{pH}\) value at which \(\mathrm{Mg}(\mathrm{OH})\), begins to precipitate from a solution containing \(0.10 \mathrm{M} \mathrm{Mg}^{+2}\) ion? Ksp of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is \(1 \times 10^{-11}\). (a) 3 (b) 6 (c) 9 (d) 11

Short Answer

Expert verified
The pH at which \(\mathrm{Mg}( ext{OH})_2\) begins to precipitate is 9.

Step by step solution

01

Understanding the Precipitation Condition

To determine when \(\mathrm{Mg}( ext{OH})_2\) begins to precipitate, we need to know the concentration of \(\text{OH}^-\) ions necessary to reach the solubility product constant (\(K_{sp}\)). The formula for \( \mathrm{Mg}( ext{OH})_2 \) dissociation is: \[ \mathrm{Mg}( ext{OH})_2(s) \rightleftharpoons \mathrm{Mg}^{2+}(aq) + 2\text{OH}^-(aq) \] This suggests the equation for the \(K_{sp}\) is: \[ K_{sp} = [\mathrm{Mg}^{2+}][\text{OH}^-]^2 \] We have \([\mathrm{Mg}^{2+}] = 0.10 \, \text{M}\) and \(K_{sp} = 1 \times 10^{-11}\).
02

Solve for Hydroxide Ion Concentration [OH-]

To find \([\text{OH}^-]\), rearrange the \(K_{sp}\) equation: \[ 1 \times 10^{-11} = 0.10 \times [\text{OH}^-]^2 \] Solving for \([\text{OH}^-]\), we have: \[ [\text{OH}^-]^2 = \frac{1 \times 10^{-11}}{0.10} = 1 \times 10^{-10} \] Taking the square root: \[ [\text{OH}^-] = \sqrt{1 \times 10^{-10}} = 1 \times 10^{-5} \text{ M} \]
03

Convert [OH-] to pH

We know \([\text{OH}^-]\) and want to find the pH. First, find pOH: \[ \text{pOH} = -\log([\text{OH}^-]) = -\log(1 \times 10^{-5}) = 5 \] Then, use the relation between pH and pOH: \[ \text{pH} = 14 - \text{pOH} = 14 - 5 = 9 \]
04

Conclusion and Answer Selection

Based on the calculation, \(\text{pH} = 9\) is the point at which \(\mathrm{Mg}( ext{OH})_2\) begins to precipitate. Thus, the corresponding answer is option (c) 9.

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Key Concepts

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

Precipitation
In chemistry, precipitation occurs when ions in a solution combine to form an insoluble compound, which then separates from the solution as a solid. This process is significant when studying solubility and reactions in aqueous solutions. Precipitation happens when the product of the reactant ion concentrations exceeds the solubility product constant ( K_{sp} ) for the compound.
  • Consider the reaction: ext{Mg}(OH)_2(s) ightleftharpoons ext{Mg}^{2+}(aq) + 2 ext{OH}^-(aq)
  • Precipitation begins when the concentration of [ ext{Mg}^{2+}] and [ ext{OH}^-] in solution reach a level where their product equals or exceeds K_{sp} .
Understanding precipitation helps predict whether a reaction will proceed by forming a solid or staying dissolved.
Solubility Product Constant (Ksp)
The solubility product constant, or K_{sp} , is an equilibrium constant for the dissolution of a sparingly soluble compound. It represents the maximum product of the ion concentrations before the compound begins to precipitate. K_{sp} provides insight into the solubility of different compounds.
  • For Mg(OH)_2 , the dissociation is representable as: Mg(OH)_2(s) ightleftharpoons ext{Mg}^{2+}(aq) + 2 ext{OH}^-(aq) .
  • Hence, K_{sp} = [ ext{Mg}^{2+}][ ext{OH}^-]^2
The K_{sp} value for Mg(OH)_2 is 1 imes 10^{-11} . A lower K_{sp} signifies lower solubility. Understanding this value helps predict when precipitation will start in varying concentrations of ions.
Hydroxide Ion Concentration (OH-)
Hydroxide ions ( ext{OH}^- ) greatly influence the solubility of metal hydroxides. By calculating the concentration of ext{OH}^- ions, you uncover information about when a compound will precipitate. The equilibrium expression for Mg(OH)_2 solubility is crucial.
  • Starting from K_{sp} = [ ext{Mg}^{2+}][ ext{OH}^-]^2 , plug in the known [ ext{Mg}^{2+}] and K_{sp} to find [ ext{OH}^-].
  • We solve for [ ext{OH}^-] : [ ext{OH}^-]^2 = rac{1 imes 10^{-11}}{0.10} = 1 imes 10^{-10} .
After taking the square root: [ ext{OH}^-] = 1 imes 10^{-5} ext{ M} . This value reveals the point at which hydroxide concentration will trigger precipitation.
pH and pOH Relationship
Understanding pH and ext{pOH} is essential in the realm of chemical equilibria. Since pH measures hydrogen ion concentration, while ext{pOH} measures hydroxide ion concentration, their relationship is inversely connected through water's constant ion product.
  • The equation for their relation is: ext{pH} + ext{pOH} = 14 .
  • When [ ext{OH}^-] = 1 imes 10^{-5} ext{ M} , calculate ext{pOH} as 5, using - ext{log}([ ext{OH}^-]) .
  • To find ext{pH} : ext{pH} = 14 - ext{pOH} = 9 .
Identifying the ext{pH} allows you to predict at what point a compound like ext{Mg}(OH)_2 starts to precipitate, offering a clear understanding of acid-base equilibria in the solution.

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Most popular questions from this chapter

A buffer solution can be prepared from a mixture of (a) \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(\mathrm{CH}_{3} \mathrm{COOH}\) in water (b) \(\mathrm{CH}_{3} \mathrm{COONa}\) and \(\mathrm{HCl}\) in water under certain conditions (c) \(\mathrm{NH}_{4} \mathrm{OH}\) and \(\mathrm{NH}_{4} \mathrm{Cl}\) in water (d) \(\mathrm{NaCl}\) and \(\mathrm{HCl}\) in water

Which of the following solutions will have no effect on \(\mathrm{pH}\) on dilution? (a) \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONa}\) (b) \(1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COONH}_{4}\) (c) \(0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{OH}+0.1 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) (d) \(0.5 \mathrm{M} \mathrm{H}_{2} \mathrm{CO}_{2}+0.5 \mathrm{M} \mathrm{NaHCO}_{3}\)

The number of \(\mathrm{H}^{+}\)ions present in \(1 \mathrm{~cm}^{3}\) of a solution whose \(\mathrm{pH}\) is 10 is (a) \(10^{-10}\) (b) \(10^{-13}\) (c) \(6.02 \times 10^{10}\) (d) \(6.02 \times 10^{13}\)

Silver acetate is a slightly soluble salt of a weak acid \(\left(\mathrm{K}_{\mathrm{a}}=1.75 \times 10^{-5}\right) .\) At \(20^{\circ} \mathrm{C}, 100 \mathrm{~g}\) of water dissolves \(1.04 \mathrm{~g}\) of crystalline silver acetate. The density of saturated solution of silver acetate at \(20^{\circ} \mathrm{C}\) is \(1.01 \mathrm{~g} / \mathrm{cc}\). The solubility product constant for silver acetate at \(20^{\circ} \mathrm{C}\) (a) \(2.43 \times 10^{-3}\) (b) \(3.87 \times 10^{-3}\) (c) \(7.74 \times 10^{-5}\) (d) \(1.35 \times 10^{-5}\)

Solubility of \(\mathrm{AgCl}\) in water, \(0.01 \mathrm{M} \mathrm{CaCl}_{2}, 0.01 \mathrm{M}\) \(\mathrm{NaCl}\) and \(0.05 \mathrm{M} \mathrm{AgNO}_{3}\) are \(\mathrm{S}_{1}, \mathrm{~S}_{2}, \mathrm{~S}_{3}\) and \(\mathrm{S}_{4}\) respec- tively then (a) \(\mathrm{S}_{1}>\mathrm{S}_{2}=\mathrm{S}_{3}>\mathrm{S}_{4}\) (b) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{4}>\mathrm{S}_{2}\) (c) \(\mathrm{S}_{1}<\mathrm{S}_{2}<\mathrm{S}_{3}<\mathrm{S}_{4}\) (d) \(\mathrm{S}_{1}>\mathrm{S}_{3}>\mathrm{S}_{2}>\mathrm{S}_{4}\)
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