Chapter 22: Problem 77
Given that \(K_{\mathrm{f}}=0.05 \mathrm{M}^{-1}\) for $$ \mathrm{Zn}(\mathrm{OH})_{2}(s)+2 \mathrm{OH}^{-}(a q) \leftrightharpoons\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}(a q) $$ calculate the solubility of \(\mathrm{Zn}(\mathrm{OH})_{2}(s)\) in a solution buffered at \(\mathrm{pH}=12.00\) at \(25^{\circ} \mathrm{C}\).
Short Answer
Step by step solution
Determine OH- Concentration from pH
Set Up Solubility Reaction
Use Formation Constant Formula
Calculate [Zn(OH)4]2-
Interpret Result as Solubility
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product
For compounds like zinc hydroxide (\(\text{Zn(OH)}_2\)), which have low solubility in water, the solubility product symbolized as (\(K_{sp}\)) can help determine how much of the compound dissociates into ions in solution. Although we didn’t directly calculate (\(K_{sp}\)) in the original exercise, understanding (\(K_{sp}\)) is helpful for grasping why solubility often depends on other factors like pH and the presence of common ions.
- The (\(K_{sp}\)) is unique for each substance and temperature.
- High pH can shift equilibrium for hydroxide-containing salts.
- Useful in predicting precipitation reactions.
Formation Constant
In our exercise, zinc hydroxide dissolves in the presence of excess hydroxide ions to form (\(\text{Zn(OH)}_4^{2-}\)), a complex ion. (\(K_f\)) indicates this reaction's favorability in forming stable complex ions, thereby increasing overall solubility compared to a situation without complexation:
- The reaction for our ion formation has (\(K_f = 0.05 \,\text{M}^{-1}\)).
- A higher (\(K_f\)) suggests more stable complexes, influencing solubility.
- The presence of ligands like (\(\text{OH}^-\)) at high concentrations can lead to a complete shift in equilibrium, favoring complex ion formation.
Hydroxide Ion Concentration
In this exercise, the solution's pH is 12, correlating to a pOH of 2, as calculated from:
(\[\text{pOH} = 14 - \text{pH} = 14 - 12 = 2\])
To find (\[\text{OH}^-\]), use:
(\[[\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} = 0.01 \,\text{M}\]).
- High hydroxide ion levels often lead to increased solubility for metal hydroxides.
- Precise measurement of (\[\text{OH}^-\]) is critical as it affects (\(K_f\)) calculations.
- Altering hydroxide concentration can drive solutions toward precipitation or increased ionic complexation.
pH and pOH Calculations
Given a pH of 12 in this problem, calculating pOH becomes straightforward using the formula above. From the given (\(\text{pH} = 12\)), we find:
(\[\text{pOH} = 14 - \text{pH} = 2\])
Using the pOH, we calculated the hydroxide ion concentration, a critical step since it impacts the solubility of zinc hydroxide by influencing the state of the solubility equilibrium.
- Calculating pH or pOH involves taking the negative logarithm of hydrogen or hydroxide ion concentration.
- These calculations are fundamental for understanding acid-base equilibria.
- Changes in pH can dramatically alter the solubility characteristics of compounds, especially those forming complexes in the solution.