`pH` of a solution of `10 ml`. `1 N` sodium acetate and `50 ml 2N` acetic acid `(K_(a)=1.8xx10^(-5))` is approximately

Options

A. `4`
B. `5`
C. `6`
D. `7`

Explanation

To find the pH of a solution containing 10 mL of 1 N sodium acetate and 50 mL of 2 N acetic acid, we can use the Henderson-Hasselbalch equation, which is given by:

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

### Step 1: Calculate pK_a

The dissociation constant \( K_a \) for acetic acid is given as \( 1.8 \times 10^{-5} \).

To find \( pK_a \):

\[ pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) \]

Using properties of logarithms:

\[ pK_a = -\log(1.8) - \log(10^{-5}) = -0.255 + 5 = 4.745 \]

### Step 2: Calculate the number of moles of the salt (sodium acetate)

The concentration of sodium acetate is 1 N, and the volume is 10 mL.

Number of moles of sodium acetate:

\[ \text{Moles of Salt} = \text{Concentration} \times \text{Volume} = 1 \, \text{N} \times 0.010 \, \text{L} = 0.010 \, \text{moles} \]

### Step 3: Calculate the number of moles of the acid (acetic acid)

The concentration of acetic acid is 2 N, and the volume is 50 mL.

Number of moles of acetic acid:

\[ \text{Moles of Acid} = \text{Concentration} \times \text{Volume} = 2 \, \text{N} \times 0.050 \, \text{L} = 0.100 \, \text{moles} \]

### Step 4: Substitute values into the Henderson-Hasselbalch equation

Now we can substitute the values into the equation:

\[ \text{pH} = pK_a + \log\left(\frac{\text{Moles of Salt}}{\text{Moles of Acid}}\right) \]

Substituting the values we calculated:

\[ \text{pH} = 4.745 + \log\left(\frac{0.010}{0.100}\right) \]

Calculating the logarithm:

\[ \log\left(\frac{0.010}{0.100}\right) = \log(0.1) = -1 \]

So, we have:

\[ \text{pH} = 4.745 - 1 = 3.745 \]

### Final Answer

The approximate pH of the solution is:

\[ \text{pH} \approx 3.75 \]