Chapter 18: Problem 64
Will a precipitate of \(\mathrm{Mg}(\mathrm{OH})_{2}\) form when \(25.0 \mathrm{mL}\) of \(0.010 \mathrm{M} \mathrm{NaOH}\) is combined with \(75.0 \mathrm{mL}\) of a \(0.10 \mathrm{M}\) solution of magnesium chloride?
Short Answer
Expert verified
Yes, a precipitate of \(\text{Mg(OH)}_2\) will form.
Step by step solution
01
Determine Concentration After Mixing
First, calculate the concentration of each ion after mixing. The total volume after mixing is \(25.0\, \text{mL} + 75.0\, \text{mL} = 100.0\, \text{mL}\) or \(0.100\, \text{L}\). The concentration of \(\text{OH}^-\) ions from \(\text{NaOH}\) is \(\frac{0.010\, \text{M} \times 0.025\, \text{L}}{0.100\, \text{L}} = 0.0025\, \text{M}\). The concentration of \(\text{Mg}^{2+}\) ions is \(\frac{0.10\, \text{M} \times 0.075\, \text{L}}{0.100\, \text{L}} = 0.075\, \text{M}\).
02
Write the Solubility Product Expression
The solubility product constant \(K_{sp}\) for \(\text{Mg(OH)}_2\) is given by the equation: \[K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2.\] The \(K_{sp}\) for \(\text{Mg(OH)}_2\) is around \(1.8 \times 10^{-11}\) at room temperature.
03
Calculate the Reaction Quotient
Using the concentrations after mixing, calculate the reaction quotient \(Q\): \[ Q = [\text{Mg}^{2+}][\text{OH}^-]^2 = (0.075)(0.0025)^2 = 4.6875 \times 10^{-7}. \]
04
Compare Q and Ksp
Since \(Q = 4.6875 \times 10^{-7}\) and \(K_{sp} = 1.8 \times 10^{-11}\), \(Q > K_{sp}\). This means that the reaction exceeds the solubility threshold, so a precipitate will form.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant
The solubility product constant, often denoted as \( K_{sp} \), is a key concept in understanding precipitation reactions in chemistry. It represents the equilibrium constant for a solid substance dissolving in water. When a substance is soluble, it will dissociate into its ions, and \( K_{sp} \) quantifies the maximum balance between these ions in a saturated solution.
The formula for \( K_{sp} \) depends on the ions produced by the solid. For magnesium hydroxide, \( \text{Mg(OH)}_2 \), the ions are \( \text{Mg}^{2+} \) and \( \text{OH}^- \). Therefore, the solubility product expression is:
The formula for \( K_{sp} \) depends on the ions produced by the solid. For magnesium hydroxide, \( \text{Mg(OH)}_2 \), the ions are \( \text{Mg}^{2+} \) and \( \text{OH}^- \). Therefore, the solubility product expression is:
- \( K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \)
Reaction Quotient
The reaction quotient, symbolized as \( Q \), is a dynamic expression used to assess the condition of a reaction at any point in time, not just at equilibrium. It is similar to the solubility product constant, but instead of indicating a saturation point, \( Q \) tells us the current state of ion concentrations in a solution.
When evaluating whether a precipitate will form, compare \( Q \) to \( K_{sp} \):
When evaluating whether a precipitate will form, compare \( Q \) to \( K_{sp} \):
- If \( Q < K_{sp} \), the solution is unsaturated, and no precipitate will form.
- If \( Q = K_{sp} \), the solution is at equilibrium, saturated with ions, but no precipitate is forming.
- If \( Q > K_{sp} \), the solution is supersaturated, and a precipitate will form.
- \( Q = [\text{Mg}^{2+}][\text{OH}^-]^2 \)
Molarity
Molarity, represented by the symbol \( M \), is the measure of the concentration of a solution. It is defined as the number of moles of solute per liter of solution and is expressed as \( \text{mol/L} \).
In many chemical calculations, including those determining whether a precipitate will form, it's crucial to calculate the molarity of ions after solutions are mixed. For instance, when mixing
For example:
In many chemical calculations, including those determining whether a precipitate will form, it's crucial to calculate the molarity of ions after solutions are mixed. For instance, when mixing
- \( 0.010 \text{ M} \text{ NaOH} \) with \( 0.10 \text{ M} \text{ MgCl}_2 \).
For example:
- Total volume after mixing: \( 100 \text{ mL} \) or \( 0.100 \text{ L} \).
- \( [\text{OH}^-] = \frac{0.010 \text{ M} \times 0.025 \text{ L}}{0.100 \text{ L}} = 0.0025 \text{ M} \)
- \( [\text{Mg}^{2+}] = \frac{0.10 \text{ M} \times 0.075 \text{ L}}{0.100 \text{ L}} = 0.075 \text{ M} \)
Precipitate Formation
Precipitate formation occurs when the product of the concentrations of the ions in solution exceeds the solubility product constant \( K_{sp} \) of the compound. When \( Q > K_{sp} \), the solution is in a supersaturated state, allowing excess ions to exit the solution as a solid precipitate.
This process is crucial in various chemical reactions and industries, from water purification to the extraction of minerals.
In the example with magnesium hydroxide, the calculated \( Q \) was:
This process is crucial in various chemical reactions and industries, from water purification to the extraction of minerals.
In the example with magnesium hydroxide, the calculated \( Q \) was:
- \( Q = 4.6875 \times 10^{-7} \)
- \( K_{sp} = 1.8 \times 10^{-11} \)