What is the pH of a solution of 0.28 M acid and 0.84 M of its conjugate base if the ionization constant of acid is `4xx10^(-4)`?

Options

A. `3.88`
B. `3.34`
C. 7
D. `10.12`

Answer

To find the pH of a solution containing 0.28 M of an acid and 0.84 M of its conjugate base, we can use the Henderson-Hasselbalch equation, which is given by:

\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]

### Step 1: Calculate pK_a
The ionization constant \( K_a \) of the acid is given as \( 4 \times 10^{-4} \). We can calculate \( pK_a \) using the formula:

\[ \text{pK}_a = -\log(K_a) \]

Substituting the value of \( K_a \):

\[ \text{pK}_a = -\log(4 \times 10^{-4}) = -\log(4) - \log(10^{-4}) = -\log(4) + 4 \]

Using a calculator, we find:

\[ \log(4) \approx 0.6 \]

Thus,

\[ \text{pK}_a \approx 4 - 0.6 = 3.4 \]

### Step 2: Apply the Henderson-Hasselbalch equation
Now we can substitute the values into the Henderson-Hasselbalch equation. We have:

- Concentration of the conjugate base = 0.84 M
- Concentration of the acid = 0.28 M

Substituting these into the equation:

\[ \text{pH} = 3.4 + \log\left(\frac{0.84}{0.28}\right) \]

### Step 3: Calculate the log term
First, calculate the ratio:

\[ \frac{0.84}{0.28} = 3 \]

Now, calculate the logarithm:

\[ \log(3) \approx 0.48 \]

### Step 4: Final calculation of pH
Now substitute this value back into the pH equation:

\[ \text{pH} = 3.4 + 0.48 = 3.88 \]

### Conclusion
Thus, the pH of the solution is approximately:

\[ \text{pH} \approx 3.88 \]