In the titration of solution of a weak acid HA and NaOH, the pH is 5.0 after 10 mL of NaOH solution has been added and 5.60 after 20 mL NaOH has been added.
What is the value of `pK_(a)` for HA?

Options

A. `5.15`
B. `5.3`
C. `5.6`
D. None of these

Answer

To find the value of \( pK_a \) for the weak acid \( HA \) based on the provided pH values during the titration with NaOH, we can follow these steps:

### Step 1: Understand the Buffer System
In the titration of a weak acid with a strong base, a buffer solution is formed. The pH of the solution can be described using the Henderson-Hasselbalch equation:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
where \( [A^-] \) is the concentration of the conjugate base and \( [HA] \) is the concentration of the weak acid.

### Step 2: Set Up the Equations
Let \( A \) be the initial moles of the weak acid \( HA \), and let \( X \) be the molarity of the NaOH solution. After adding 10 mL of NaOH, the pH is 5.0:
\[ 5.0 = pK_a + \log\left(\frac{10X}{A - 10X}\right) \tag{1} \]
After adding 20 mL of NaOH, the pH is 5.60:
\[ 5.60 = pK_a + \log\left(\frac{20X}{A - 20X}\right) \tag{2} \]

### Step 3: Subtract the Two Equations
Subtract equation (1) from equation (2):
\[ (5.60 - 5.0) = \log\left(\frac{20X}{A - 20X}\right) - \log\left(\frac{10X}{A - 10X}\right) \]
This simplifies to:
\[ 0.60 = \log\left(\frac{20X}{A - 20X} \cdot \frac{A - 10X}{10X}\right) \]
Using the property of logarithms, we can rewrite this as:
\[ 0.60 = \log\left(\frac{20(A - 10X)}{10(A - 20X)}\right) \]

### Step 4: Exponentiate to Remove the Logarithm
Exponentiating both sides gives:
\[ 10^{0.60} = \frac{20(A - 10X)}{10(A - 20X)} \]
Calculating \( 10^{0.60} \) gives approximately \( 4 \):
\[ 4 = \frac{20(A - 10X)}{10(A - 20X)} \]
This simplifies to:
\[ 4(A - 20X) = 2(A - 10X) \]

### Step 5: Solve for \( A \)
Expanding both sides:
\[ 4A - 80X = 2A - 20X \]
Rearranging gives:
\[ 2A = 60X \implies A = 30X \]

### Step 6: Substitute Back to Find \( pK_a \)
Substituting \( A = 30X \) back into equation (1):
\[ 5.0 = pK_a + \log\left(\frac{10X}{30X - 10X}\right) \]
This simplifies to:
\[ 5.0 = pK_a + \log\left(\frac{10}{20}\right) = pK_a + \log\left(\frac{1}{2}\right) \]
Thus:
\[ 5.0 = pK_a - 0.301 \]
Solving for \( pK_a \):
\[ pK_a = 5.0 + 0.301 = 5.301 \]

### Final Answer
The value of \( pK_a \) for the weak acid \( HA \) is approximately **5.30**.

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