Question

Acetic acid has a \(K_{a}\) of \(1.8 \times 10^{-5},\) and ammonia has a \(K_{b}\) of \(1.8 \times 10^{-5} .\) Find \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right],\left[\mathrm{OH}^{-}\right], \mathrm{pH},\) and \(\mathrm{pOH}\) for (a) \(0.240 \mathrm{M}\) acctic acid and (b) \(0.240 \mathrm{M}\) ammonia.

Short answer

For (a) Acetic Acid: [\text{H}_3\text{O}^+] = 2.08 \times 10^{-3}\text{M}, [\text{OH}^-] = 4.81 \times 10^{-12}\text{M}, pH = 2.68, pOH = 11.32. For (b) Ammonia: [\text{OH}^-] = 2.08 \times 10^{-3}\text{M}, pOH = 2.68, [\text{H}_3\text{O}^+] = 4.81 \times 10^{-12}\text{M}, pH = 11.32.

Step-by-step solution

Calculate \(\text{[H}_3\text{O}^+\)\text{ concentration for Acetic Acid}

For acetic acid (CH_3COOH), use the expression for the ionization constant \(K_a\) to set up the equation: \[ K_a = \frac{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]Given: \(K_a = 1.8 \times 10^{-5}\) and initial concentration of acetic acid \(0.240 \text{M}\). Assume \( [\text{H}_3\text{O}^+] = x \). Substitute values and solve for \(x\): \[ 1.8 \times 10^{-5} = \frac{x\cdot x}{0.240 - x} \] Approximating \(0.240 - x \approx 0.240\): \[ 1.8 \times 10^{-5} = \frac{x^2}{0.240} \] Thus, \[ x^2 = 1.8 \times 10^{-5} \times 0.240 \] \[ x^2 = 4.32 \times 10^{-6} \]\[ x = \sqrt{4.32 \times 10^{-6}} \] \[ x = 2.08 \times 10^{-3} \text{M}\] So, \( [\text{H}_3\text{O}^+] = 2.08 \times 10^{-3} \text{M}\)

Calculate the pH for Acetic Acid

Use the expression: \[ \text{pH} = -\log [\text{H}_3\text{O}^+] \] Given: \( [\text{H}_3\text{O}^+] = 2.08 \times 10^{-3}\): \[ \text{pH} = -\log (2.08 \times 10^{-3}) \] \[ \text{pH} \approx 2.68 \]

Calculate the concentration of \( \text{OH}^-\) for Acetic Acid

Use the relationship between \( [\text{H}_3\text{O}^+]\) and \( [\text{OH}^-]\), which is: \[ [\text{H}_3\text{O}^+][\text{OH}^-] = 1 \times 10^{-14} \] Given: \( [\text{H}_3\text{O}^+] = 2.08 \times 10^{-3}\): \[ (2.08 \times 10^{-3})[\text{OH}^-] = 1 \times 10^{-14} \] \[ [\text{OH}^-] = \frac{1 \times 10^{-14}}{2.08 \times 10^{-3}} \] \[ [\text{OH}^- ]= 4.81 \times 10^{-12} \text{M} \]

Calculate the pOH for Acetic Acid

Use the expression: \[ \text{pOH} = -\log [\text{OH}^-] \] Given: \( [\text{OH}^-] = 4.81 \times 10^{-12}\): \[ \text{pOH} = -\log (4.81 \times 10^{-12}) \] \[ \text{pOH} \approx 11.32 \]

Calculate \( [\text{OH}^-]\) concentration for Ammonia

For ammonia (NH_3), use the expression for the ionization constant \(K_b\) to set up the equation: \[ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} \] Given: \( K_b = 1.8 \times 10^{-5}\), initial concentration \( [NH_3] = 0.240 \text{M}\). Assume \[ [\text{OH}^-] = x \]. Substitute values and solve for \( x \)\[ 1.8 \times 10^{-5} = \frac{x \cdot x}{0.240} \] \[ x^2 = 1.8 \times 10^{-5} \times 0.240 \] \[ x^2 = 4.32 \times 10^{-6} \] \[ x = \sqrt{4.32 \times 10^{-6}} \] \[ x = 2.08 \times 10^{-3}\text{M} \] So, \( [\text{OH}^-] = 2.08 \times 10^{-3} \text{M}\)

Calculate the \( \text{pOH}\) for Ammonia

Use the expression: \[ \text{pOH} = -\log [\text{OH}^-] \]Given: \( [\text{OH}^-] = 2.08 \times 10^{-3} \) \[ \text{pOH} = -\log (2.08 \times 10^{-3}) \] \[ \text{pOH} = 2.68 \]

Calculate the \( [\text{H}_3\text{O}^+]\) concentration for Ammonia

Use the relationship between \( [\text{H}_3\text{O}^+] \) and \( [\text{OH}^-]\), which is: \[ [\text{H}_3\text{O}^+] [\text{OH}^-] = 1 \times 10^{-14} \]Given: \( [\text{OH}^-] = 2.08 \times 10^{-3}\): \[ (\text{2.08} \times 10^{-3})[\text{H}_3\text{O}^+] = 1 \times 10^{-14} \] \[ [\text{H}_3\text{O}^+] = \frac{1 \times 10^{-14}}{2.08 \times 10^{-3}} \] \[ [\text{H}_3\text{O}^+] = 4.81 \times 10^{-12}\text{M} \]

Calculate the \( \text{pH}\) for Ammonia

Use the expression: \[ \text{pH} = -\log [\text{H}_3\text{O}^+] \] Given: \( [\text{H}_3\text{O}^+] = 4.81 \times 10^{-12} \): \[ \text{pH} = -\log (4.81 \times 10^{-12}) \] \[ \text{pH} = 11.32 \]

Key concepts

Acetic Acid Ionization

Understanding the ionization of acetic acid is essential in acid-base equilibrium calculations.
Acetic acid (\text{CH}_3\text{COOH}) is a weak acid, and its ionization in water is represented by the equation: \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}_3\text{O}^+.
The ionization constant (\text{K}_\text{a}) for acetic acid is given by:
\[ \text{K}_\text{a} = \frac{{[\text{H}_3\text{O}^+][\text{CH}_3\text{COO}^-]}}{{[\text{CH}_3\text{COOH}]}} \]
In this formula, the concentration of \text{H}_3\text{O}^+ and \text{CH}_3\text{COO}^- are products, while \text{CH}_3\text{COOH} is the reactant.
If you're given an initial acetic acid concentration (\text{C}_0), you can make the assumption \text{[H}_3\text{O}^+] = x.
Substituting the values into the expression and solving for x yields the concentration of \text{H}_3\text{O}^+.
This follows the steps:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.240 - x} \]
Approximating that 0.240 - x is approximately 0.240 simplifies the math to:
\[ x = \text{{}}2.08 \times 10^{-3} \text{M} \]
This calculation helps you find the concentration of \text{H}_3\text{O}^+ in the solution.

Ammonia Ionization

Ammonia (\text{NH}_3) is a weak base and its ionization in water can be expressed with the following equation: \text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-.
The ionization constant (\text{K}_\text{b}) for ammonia can be calculated using:
\[ \text{K}_\text{b} = \frac{{[\text{NH}_4^+][\text{OH}^-]}}{{[\text{NH}_3]}} \]
Given an initial concentration of ammonia (\text{C}_0), assume the approximate \text{[OH}^-] concentration is (x).
Substituting the values and solving for x gives the concentration of hydroxide ions in the solution:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.240}\]
\[ x = \text{{}}2.08 \times 10^{-3} \text{M} \]
Thus, the concentration of \text{OH}^- can be calculated.
This process helps in understanding how weak bases ionize in water and their resulting equilibrium concentrations.

pH and pOH Calculations

Calculating pH and pOH is fundamental in determining the acidity or basicity of a solution.
The pH of a solution is calculated using the formula:
\[ \text{{pH}} = -\text{{log}}(\text{{[H}_3\text{{O}}^+]})\]
For example, for \text{{[H}_3\text{{O}}^+]} = 2.08 \times 10^{-3}, the pH will be:
\[ \text{{pH}} = -\text{{log}}(2.08 \times 10^{-3}) \]
This calculation results in a pH of approximately 2.68.
Similarly, the pOH of a solution is calculated using:
\[ \text{{pOH}} = -\text{{log}}(\text{{[OH}^-]})\]
For \text{{[OH}^-]} = 2.08 \times 10^{-3}, the pOH will be:
\[ \text{{pOH}} = -\text{{log}}(2.08 \times 10^{-3}) \]
This gives a pOH of about 2.68.
pH and pOH are linked by the relation:
\[ \text{{pH}} + \text{{pOH}} = 14 \]
Therefore, knowing either pH or pOH allows you to calculate the other.
This is crucial for solving problems involving acid-base equilibria.

Ionization Constants (Ka and Kb)

Ionization constants are key in understanding the strength of acids and bases.
\text{K}_\text{a} measures the ionization of a weak acid in water.
A high \text{K}_\text{a} value indicates a strong tendency to lose a proton.\text{K}_\text{b} measures the ionization of a weak base in water.
A high \text{K}_\text{b} value signifies a strong ability to gain a proton.
For example, the \text{K}_\text{a} value for acetic acid is 1.8 \times 10^{-5}\text{}, meaning it partially ionizes in water.
Similarly, the \text{K}_\text{b} value for ammonia is 1.8 \times 10^{-5}\text{}, indicating its partial ionization.
\text{K}_\text{a} and \text{K}_\text{b} are related to the ionization of water via:\br> \[ \text{K}_\text{w} = \text{K}_\text{a} \times \text{K}_\text{b} \]
where \text{K}_\text{w} is the ionization constant of water, equal to 1.0 \times 10^{-14} at 25°C.
Understanding these constants aids in predicting the behavior of acids and bases in solutions.