Question

Strong base is dissolved in 665 ml. of 0.200 M weak acid (K, 4.73 x 10-5) to make a buffer with a pH of 4.10, Assume that the volume remains constant when the base is added.

HA(aq) + OH (aq)

H₂O(l) + A(aq)

How many moles of strong base were initially added?

Answer 1

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.