To solve this problem, we can use the Henderson-Hasselbalch equation to relate the pH of the buffer solution to the concentrations of the weak acid (HA) and its conjugate base (A⁻).
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Step 1: Determine \( \text{pKa} \)
First, we need to find the \( \text{pKa} \) from the given \( K_a \) value:
\[ K_a = 4.73 \times 10^{-5} \]
We calculate \( \text{pKa} \) using the relationship:
\[ \text{pKa} = -\log(K_a) \]
Substituting the value:
\[ \text{pKa} = -\log(4.73 \times 10^{-5}) \approx 4.33 \]
Step 2: Set up the equation with the given pH
We know the pH of the buffer solution:
\[ \text{pH} = 4.10 \]
Using the Henderson-Hasselbalch equation:
\[ 4.10 = 4.33 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Step 3: Solve for the ratio \(\frac{[\text{A}^-]}{[\text{HA}]}\)
Rearranging the equation gives:
\[ \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 4.10 - 4.33 = -0.23 \]
Converting from logarithmic form to ratio form:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{-0.23} \approx 0.585 \]
Step 4: Calculate moles of weak acid (HA)
The volume of the weak acid solution is 665 mL, which we convert to liters:
\[ V = 665 , \text{mL} = 0.665 , \text{L} \]
Now we calculate the moles of the weak acid (\([\text{HA}]\)) in the solution:
\[ [\text{HA}] = 0.200 , \text{M} \times 0.665 , \text{L} = 0.133 , \text{moles} \]
Step 5: Set up the relationship between moles of HA and A⁻
Let the moles of the weak acid (HA) be \( n_{HA} = 0.133 \) moles. Let the moles of the strong base added be \( n_{OH^-} \). When OH⁻ is added, it reacts with HA to produce A⁻:
\[ \text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O} \]
After reaction:
- Moles of A⁻ formed = \( n_{OH^-} \)
- Moles of HA remaining = \( n_{HA} - n_{OH^-} \)
Using the ratio calculated above:
\[ \frac{n_{OH^-}}{n_{HA} - n_{OH^-}} = 0.585 \]
Step 6: Substitute and solve for \( n_{OH^-} \)
Let \( n_{OH^-} = x \):
\[ \frac{x}{0.133 - x} = 0.585 \]
Cross-multiplying gives:
\[ x = 0.585 (0.133 - x) \]
Expanding:
\[ x = 0.078705 - 0.585x \]
Combining like terms:
\[ x + 0.585x = 0.078705 \]
\[ 1.585x = 0.078705 \]
Solving for \( x \):
\[ x = \frac{0.078705}{1.585} \approx 0.0496 , \text{moles} \]
Final Answer
Thus, the number of moles of strong base (OH⁻) initially added is approximately \( 0.0496 \) moles.
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