The ratio of `pH` of solution `(1)` containing 1 mole of` CH_(3)COONa` and 1 mole of `HCl` and solution `(II)` containing 1 mole of `CH_(3)COONa` and 1 mole of acetic acid in one litre is `:`

Options

A. `1:2`
B. `2:1`
C. `1:3`
D. `3:1`

Explanation

To solve the problem, we need to find the ratio of the pH of two solutions:

1. Solution (I): Contains 1 mole of sodium acetate (CH₃COONa) and 1 mole of hydrochloric acid (HCl).
2. Solution (II): Contains 1 mole of sodium acetate (CH₃COONa) and 1 mole of acetic acid (CH₃COOH).

### Step-by-Step Solution:

**Step 1: Analyze Solution (I)**

In Solution (I), sodium acetate reacts with hydrochloric acid:
\[ \text{CH}_3\text{COONa} + \text{HCl} \rightarrow \text{CH}_3\text{COOH} + \text{NaCl} \]

- Initially, we have 1 mole of sodium acetate and 1 mole of HCl.
- After the reaction, we will have 1 mole of acetic acid (CH₃COOH) and 1 mole of sodium chloride (NaCl) in the solution.

**Step 2: Determine the pH of Solution (I)**

Acetic acid is a weak acid and will dissociate partially in the solution:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

Let \( K_a \) be the dissociation constant of acetic acid. The concentration of acetic acid is 1 M (since we have 1 mole in 1 liter).

Using the formula for the concentration of hydrogen ions (\( [H^+] \)):
\[ [H^+] = \sqrt{K_a \cdot C} \]
Where \( C \) is the concentration of acetic acid.

Since \( C = 1 \):
\[ [H^+] = \sqrt{K_a} \]

Now, the pH of Solution (I) can be calculated as:
\[ \text{pH}_1 = -\log[H^+] = -\log(\sqrt{K_a}) = -\frac{1}{2} \log(K_a) \]

This can be rewritten as:
\[ \text{pH}_1 = \frac{1}{2} \text{pK}_a \]

**Step 3: Analyze Solution (II)**

In Solution (II), we have a buffer solution made of sodium acetate and acetic acid:
- 1 mole of sodium acetate (CH₃COONa) and 1 mole of acetic acid (CH₃COOH) in 1 liter.

Using the Henderson-Hasselbalch equation for buffer solutions:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Here, both the salt (sodium acetate) and the acid (acetic acid) are present in equal concentrations (1 M):
\[ \text{pH}_2 = \text{pK}_a + \log(1) \]
Since \( \log(1) = 0 \):
\[ \text{pH}_2 = \text{pK}_a \]

**Step 4: Find the Ratio of pH Values**

Now we have:
- \( \text{pH}_1 = \frac{1}{2} \text{pK}_a \)
- \( \text{pH}_2 = \text{pK}_a \)

To find the ratio of pH values:
\[ \frac{\text{pH}_1}{\text{pH}_2} = \frac{\frac{1}{2} \text{pK}_a}{\text{pK}_a} = \frac{1}{2} \]

### Final Answer:
The ratio of pH of Solution (I) to Solution (II) is:
\[ \text{pH}_1 : \text{pH}_2 = 1 : 2 \]