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

Write the hydrolysis reaction for Al(H₂O)₆³⁺

The aluminum ion, Al(H₂O)₆³⁺, will react with water to form hydronium (H₃O⁺) ions and the complex Al(H₂O)₅(OH)²⁺: \[ \ce{Al(H2O)6^{3+} + H2O <=> Al(H2O)5(OH)^{2+} + H3O^{+}} \]

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 × 10⁻⁵: \[ K_{a} = \frac{[\ce{Al(H2O)5(OH)^{2+}}][\ce{H3O^{+}}]}{[\ce{Al(H2O)6^{3+}}]} \]

Calculate the initial concentrations

Since the initial concentration of Al(NO₃)₃ is 0.050 M, the initial concentration of Al(H₂O)₆³⁺ ions in the solution is also 0.050 M. The initial concentrations of Al(H₂O)₅(OH)²⁺ ions and H₃O⁺ ions are zero: Initial concentrations: \[ [\ce{Al(H2O)6^{3+}}]_{0} = 0.050\,\mathrm{M} \] \[ [\ce{Al(H2O)5(OH)^{2+}}]_{0} = 0\,\mathrm{M} \] \[ [\ce{H3O^{+}}]_{0} = 0\,\mathrm{M} \]

Determine the change in concentrations

Define x as the moles of Al(H₂O)₅(OH)²⁺ and H₃O⁺ ions formed during the hydrolysis reaction. Then the change in concentrations can be given by: \[ [\ce{Al(H2O)5(OH)^{2+}}] = +x \] \[ [\ce{H3O^{+}}] = +x \] \[ [\ce{Al(H2O)6^{3+}}] = -x \]

Calculate the final concentrations

The final concentrations are formed after the hydrolysis reaction takes place, and we can express them as follows: \[ [\ce{Al(H2O)5(OH)^{2+}}] = x \] \[ [\ce{H3O^{+}}] = x \] \[ [\ce{Al(H2O)6^{3+}}] = 0.050 - x \]

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: \[ 1.4 \times 10^{-5} \approx \frac{x^2}{0.050} \]

Solve for x

Now we can solve for x, which is the concentration of H₃O⁺ ions in the solution: \[ x^2 = 1.4 \times 10^{-5} \times 0.050 \] \[ x = \sqrt{7 \times 10^{-7}} \] \[ x = 2.65 \times 10^{-4}\,\mathrm{M} \]

Calculate the pH

Now we can use the concentration of H₃O⁺ 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(NO₃)₃ solution is approximately 3.58.

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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

Step by step solution

Write the hydrolysis reaction for Al(H₂O)₆³⁺

Write the expression for the Ka

Calculate the initial concentrations

Determine the change in concentrations

Calculate the final concentrations

Substitute concentrations into the Ka expression

Solve for x

Calculate the pH

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