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Magazine Chapter 14: Problem 127
Calculate the \(\mathrm{pH}\) of a \(0.050-M \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.4 \times 10^{-5} .\)
Short Answer
Expert verified
The pH of a 0.050 M $\ce{Al(NO3)3}$ solution is approximately 3.58.
Step by step solution
01
Write the hydrolysis reaction for Al(H鈧侽)鈧喡斥伜
The aluminum ion, Al(H鈧侽)鈧喡斥伜, will react with water to form hydronium (H鈧僌鈦�) ions and the complex Al(H鈧侽)鈧�(OH)虏鈦�:
\[ \ce{Al(H2O)6^{3+} + H2O <=> Al(H2O)5(OH)^{2+} + H3O^{+}} \]
02
Write the expression for the Ka
Now, let's write the expression for the Ka of the hydrolysis reaction, which is given as 1.4 脳 10鈦烩伒:
\[ K_{a} = \frac{[\ce{Al(H2O)5(OH)^{2+}}][\ce{H3O^{+}}]}{[\ce{Al(H2O)6^{3+}}]} \]
03
Calculate the initial concentrations
Since the initial concentration of Al(NO鈧�)鈧� is 0.050 M, the initial concentration of Al(H鈧侽)鈧喡斥伜 ions in the solution is also 0.050 M. The initial concentrations of Al(H鈧侽)鈧�(OH)虏鈦� ions and H鈧僌鈦� ions are zero:
Initial concentrations:
\[ [\ce{Al(H2O)6^{3+}}]_{0} = 0.050\,\mathrm{M} \]
\[ [\ce{Al(H2O)5(OH)^{2+}}]_{0} = 0\,\mathrm{M} \]
\[ [\ce{H3O^{+}}]_{0} = 0\,\mathrm{M} \]
04
Determine the change in concentrations
Define x as the moles of Al(H鈧侽)鈧�(OH)虏鈦� and H鈧僌鈦� ions formed during the hydrolysis reaction. Then the change in concentrations can be given by:
\[ [\ce{Al(H2O)5(OH)^{2+}}] = +x \]
\[ [\ce{H3O^{+}}] = +x \]
\[ [\ce{Al(H2O)6^{3+}}] = -x \]
05
Calculate the final concentrations
The final concentrations are formed after the hydrolysis reaction takes place, and we can express them as follows:
\[ [\ce{Al(H2O)5(OH)^{2+}}] = x \]
\[ [\ce{H3O^{+}}] = x \]
\[ [\ce{Al(H2O)6^{3+}}] = 0.050 - x \]
06
Substitute concentrations into the Ka expression
Now, we will substitute the final concentrations into the Ka expression and solve for x:
\[ 1.4 \times 10^{-5} = \frac{x \times x}{0.050 - x} \]
As the value of Ka is quite small, we can assume x << 0.050, which leads to the following approximation:
\[ 1.4 \times 10^{-5} \approx \frac{x^2}{0.050} \]
07
Solve for x
Now we can solve for x, which is the concentration of H鈧僌鈦� ions in the solution:
\[ x^2 = 1.4 \times 10^{-5} \times 0.050 \]
\[ x = \sqrt{7 \times 10^{-7}} \]
\[ x = 2.65 \times 10^{-4}\,\mathrm{M} \]
08
Calculate the pH
Now we can use the concentration of H鈧僌鈦� ions to calculate the pH of the solution using the pH formula:
\[ \mathrm{pH} = -\log_{10}[\ce{H3O^{+}}] \]
\[ \mathrm{pH} = -\log_{10}(2.65 \times 10^{-4}) \]
\[ \mathrm{pH} \approx 3.58 \]
So, the pH of the 0.050 M Al(NO鈧�)鈧� solution is approximately 3.58.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Acid-Base Equilibrium
Acid-base equilibrium refers to the balance between the concentrations of acids and bases in a solution. This equilibrium determines the pH of the solution. It involves the donation and acceptance of protons (H鈦� ions) between chemical species. The hydrogen ion (
H鈧僌鈦�
) concentration is crucial for calculating pH.
- An acid releases protons, increasing H鈧僌鈦� and lowering pH.
- A base accepts protons, decreasing H鈧僌鈦� and raising pH.
Hydrolysis
Hydrolysis is a reaction where water is used to break bonds in molecules. In acid-base chemistry, it describes how ions react with water to either donate or accept a proton. This process is essential for understanding the behavior of salts in water.
In the hydrolysis of Al(H鈧侽)鈧喡斥伜 , the ion reacts with water, leading to the formation of H鈧僌鈦� . This occurs because the metal ion, like aluminum, can polarize water molecules, making it easier to release a hydrogen ion. Hydrolysis shifts the acid-base equilibrium and affects the pH of the solution.
In the hydrolysis of Al(H鈧侽)鈧喡斥伜 , the ion reacts with water, leading to the formation of H鈧僌鈦� . This occurs because the metal ion, like aluminum, can polarize water molecules, making it easier to release a hydrogen ion. Hydrolysis shifts the acid-base equilibrium and affects the pH of the solution.
Ka (Acid Dissociation Constant)
The acid dissociation constant, K_a, measures the strength of an acid in solution. It indicates the extent to which an acid donates protons to water, forming hydronium ions. A larger K_a signifies a stronger acid that dissociates more completely.
The calculation of K_a involves concentrations from the equilibrium expression:
\[ K_{a} = \frac{[Al(H鈧侽)鈧�(OH)^{2+}][H鈧僌^{+}]}{[Al(H鈧侽)鈧哵{3+}]} \]
By substituting equilibrium concentrations into this formula, you can find the dissociation constant, which helps assess the acid鈥檚 impact on solution pH.
The calculation of K_a involves concentrations from the equilibrium expression:
\[ K_{a} = \frac{[Al(H鈧侽)鈧�(OH)^{2+}][H鈧僌^{+}]}{[Al(H鈧侽)鈧哵{3+}]} \]
By substituting equilibrium concentrations into this formula, you can find the dissociation constant, which helps assess the acid鈥檚 impact on solution pH.
Al(NO3)3 Solution
Al(NO鈧�)鈧�, or aluminum nitrate, dissolves in water to produce aluminum ions
(Al(H鈧侽)鈧喡斥伜)
. These ions can undergo hydrolysis, affecting the acidity of the solution. The initial concentration of 0.050 M refers to the starting amount of aluminum ions before hydrolysis begins.
This solution's behavior is influenced by:
This solution's behavior is influenced by:
- The hydrolysis of Al鲁鈦�, releasing H鈧僌鈦�, which lowers pH.
- The small K_a value, indicating incomplete dissociation and a slight acidic nature.
