\( K_a \) for \( \text{CH}_3 \text{COOH} \) is \( 1.8 \times 10^{-5} \) and \( K_b \) for \( \text{NH}_4 \text{OH} \) is \( 1.8 \times 10^{-5} \). The pH of ammonium acetate solution will be:
Correct Answer: 7
Approach Solution - 1
The pH is given by:
\(\text{pH} = \frac{pK_a + pK_b}{2}\)
Since \(pK_a = pK_b\),
\(\text{pH} = \frac{7 + 7}{2} = 7\)
Approach Solution -2
The problem asks for the pH of an ammonium acetate (\(\text{CH}_3\text{COONH}_4\)) solution, given the acid dissociation constant (\(K_a\)) for acetic acid and the base dissociation constant (\(K_b\)) for ammonium hydroxide.
Concept Used:
Ammonium acetate is a salt formed from the reaction of a weak acid (\(\text{CH}_3\text{COOH}\)) and a weak base (\(\text{NH}_4\text{OH}\)). When such a salt dissolves in water, both the cation and the anion undergo hydrolysis.
The pH of a solution of a salt of a weak acid and a weak base is independent of the salt's concentration and is given by the formula:
\[ \text{pH} = 7 + \frac{1}{2} (\text{p}K_a - \text{p}K_b) \]where:
- \( \text{p}K_a = -\log_{10}(K_a) \)
- \( \text{p}K_b = -\log_{10}(K_b) \)
Step-by-Step Solution:
Step 1: Identify the given values.
The acid dissociation constant for acetic acid (\(\text{CH}_3\text{COOH}\)) is:
\[ K_a = 1.8 \times 10^{-5} \]The base dissociation constant for ammonium hydroxide (\(\text{NH}_4\text{OH}\)) is:
\[ K_b = 1.8 \times 10^{-5} \]Step 2: Compare the values of \(K_a\) and \(K_b\).
From the given data, it is clear that the value of the acid dissociation constant is equal to the value of the base dissociation constant.
\[ K_a = K_b \]Step 3: Determine the relationship between \(\text{p}K_a\) and \(\text{p}K_b\).
Taking the negative logarithm of both sides of the equation \(K_a = K_b\):
\[ -\log_{10}(K_a) = -\log_{10}(K_b) \]By definition, this means:
\[ \text{p}K_a = \text{p}K_b \]Therefore, the difference between them is zero:
\[ \text{p}K_a - \text{p}K_b = 0 \]Step 4: Calculate the pH of the solution using the formula.
Substitute the result from Step 3 into the pH formula for a salt of a weak acid and a weak base:
\[ \text{pH} = 7 + \frac{1}{2} (\text{p}K_a - \text{p}K_b) \] \[ \text{pH} = 7 + \frac{1}{2} (0) \] \[ \text{pH} = 7 + 0 \]Final Computation & Result:
The final calculation gives:
\[ \text{pH} = 7 \]Since the acid and base strengths are equal (\(K_a = K_b\)), the hydrolysis of the cation and anion produce equal amounts of \(H^+\) and \(OH^-\) ions, respectively, resulting in a neutral solution.
The pH of the ammonium acetate solution will be 7.
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