Understanding Solubility Equilibrium
Solubility equilibrium occurs when the rate of dissolution of a solute equals the rate of precipitation of that solute in a saturated solution. At this point, the concentrations of the solute in solution remain constant. The equilibrium can be described by the following general equation:
\[
\text{Solid} \rightleftharpoons \text{Dissolved ions}
\]
For a salt such as \( \text{AB} \), which dissociates into \( \text{A}^+ \) and \( \text{B}^- \):
\[
\text{AB (s)} \rightleftharpoons \text{A}^+ (aq) + \text{B}^- (aq)
\]
The solubility product constant, \( K_{sp} \), is a key concept in solubility equilibria. It represents the product of the molar concentrations of the dissolved ions, each raised to the power of their coefficients in the balanced equation:
\[
K_{sp} = [\text{A}^+]^m \cdot [\text{B}^-]^n
\]
Where \( m \) and \( n \) are the stoichiometric coefficients of the ions in the dissolution equation.
Factors Affecting Solubility
Several factors influence the solubility of a substance:
1. Temperature
- Generally, the solubility of solids in liquids increases with temperature.
- The solubility of gases in liquids typically decreases with increasing temperature.
2. Pressure
- Pressure has a more pronounced effect on the solubility of gases.
- An increase in pressure increases the solubility of a gas in a liquid (Henry's Law).
3. Common Ion Effect
- The presence of a common ion decreases the solubility of a salt due to Le Chatelier's Principle.
- For example, adding sodium chloride (NaCl) to a saturated solution of silver chloride (AgCl) will reduce the solubility of AgCl.
4. pH of the Solution
- The solubility of some salts is influenced by the pH of the solution.
- For example, metal hydroxides are more soluble in acidic solutions.
Practice Problems on Solubility Equilibrium
To solidify your understanding of solubility equilibria, let’s work through some practice problems.
Problem 1: Finding Ksp
Given the dissolution of barium sulfate (\( \text{BaSO}_4 \)):
\[
\text{BaSO}_4 (s) \rightleftharpoons \text{Ba}^{2+} (aq) + \text{SO}_4^{2-} (aq)
\]
If the concentration of \( \text{Ba}^{2+} \) in a saturated solution is \( 0.0020 \, \text{mol/L} \), calculate the \( K_{sp} \) for barium sulfate.
Solution 1:
From the dissolution equation, we can express the \( K_{sp} \):
\[
K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]
\]
Let \( x = [\text{Ba}^{2+}] = 0.0020 \, \text{mol/L} \).
Since \( \text{BaSO}_4 \) dissociates into equal amounts of \( \text{Ba}^{2+} \) and \( \text{SO}_4^{2-} \):
\[
[\text{SO}_4^{2-}] = 0.0020 \, \text{mol/L}
\]
Thus,
\[
K_{sp} = (0.0020)(0.0020) = 4.00 \times 10^{-6}
\]
Problem 2: Common Ion Effect
Calculate the solubility of \( \text{AgCl} \) in a solution that is \( 0.10 \, \text{mol/L} \) in \( \text{NaCl} \). The \( K_{sp} \) of \( \text{AgCl} \) is \( 1.77 \times 10^{-10} \).
Solution 2:
The dissolution of silver chloride is represented as:
\[
\text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq)
\]
The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+][\text{Cl}^-]
\]
Given that \( [\text{Cl}^-] = 0.10 \, \text{mol/L} \), we can substitute this into the \( K_{sp} \) expression:
\[
1.77 \times 10^{-10} = [\text{Ag}^+](0.10)
\]
Solving for \( [\text{Ag}^+] \):
\[
[\text{Ag}^+] = \frac{1.77 \times 10^{-10}}{0.10} = 1.77 \times 10^{-9} \, \text{mol/L}
\]
The solubility (\( s \)) of \( \text{AgCl} \) in the presence of \( \text{NaCl} \) is \( 1.77 \times 10^{-9} \, \text{mol/L} \).
Problem 3: Temperature and Solubility
A salt dissolves in water with a \( K_{sp} \) of \( 3.0 \times 10^{-5} \) at \( 25^\circ C \). What will happen to the solubility of this salt if the temperature is increased to \( 50^\circ C \), assuming the \( K_{sp} \) increases to \( 5.0 \times 10^{-5} \)?
Solution 3:
At \( 25^\circ C \):
If \( s_1 \) is the solubility at \( 25^\circ C \):
\[
K_{sp} = s_1^2 = 3.0 \times 10^{-5}
\]
Thus,
\[
s_1 = \sqrt{3.0 \times 10^{-5}} \approx 5.48 \times 10^{-3} \, \text{mol/L}
\]
At \( 50^\circ C \):
If \( s_2 \) is the solubility at \( 50^\circ C \):
\[
K_{sp} = s_2^2 = 5.0 \times 10^{-5}
\]
Thus,
\[
s_2 = \sqrt{5.0 \times 10^{-5}} \approx 7.07 \times 10^{-3} \, \text{mol/L}
\]
The solubility of the salt increases from \( 5.48 \times 10^{-3} \, \text{mol/L} \) to \( 7.07 \times 10^{-3} \, \text{mol/L} \) when the temperature is raised, indicating that the salt is more soluble at higher temperatures.
Conclusion
Understanding solubility equilibrium practice problems is vital for students and professionals in chemistry. Mastery of these concepts allows for better predictions about how substances behave in various chemical environments. By working through the principles of solubility products, factors affecting solubility, and applying practical problems, one can gain a deeper appreciation of chemical equilibria and their significance in real-world applications. Engaging with these practice problems not only solidifies theoretical knowledge but also enhances analytical skills crucial for success in the field of chemistry.
Frequently Asked Questions
What is solubility equilibrium?
Solubility equilibrium is the state in which the rate of dissolution of a solute equals the rate of precipitation, resulting in a constant concentration of the solute in the solution.
How do you calculate the solubility product constant (Ksp) for a sparingly soluble salt?
To calculate Ksp, you write the equilibrium expression based on the balanced dissolution reaction, and then substitute the equilibrium concentrations of the ions into the expression.
What factors can affect the solubility of a salt in water?
Factors that affect solubility include temperature, pressure (for gases), the presence of common ions, and the pH of the solution.
Can you provide a practice problem involving Ksp and its solution?
Sure! If the Ksp of AgCl is 1.77 x 10^-10, find the solubility of AgCl in mol/L. Set up the equation: Ksp = [Ag+][Cl-] = s^2, where s is the solubility. Solve for s: s = √(1.77 x 10^-10) = 1.33 x 10^-5 mol/L.
What is the common ion effect in solubility equilibrium?
The common ion effect refers to the decrease in solubility of a salt when a common ion is added to the solution, shifting the equilibrium position according to Le Chatelier's principle.
How can you determine if a precipitate will form when two solutions are mixed?
To determine if a precipitate will form, calculate the ion product (Q) by multiplying the concentrations of the ions in solution. If Q > Ksp, a precipitate will form; if Q < Ksp, no precipitate will form.
What role does temperature play in the solubility of salts?
Temperature can significantly affect solubility; for most salts, solubility increases with temperature, while for some salts, it may decrease. Always check specific salt behavior when considering temperature effects.
