The `pH` of a solution containing `0.1mol` of `CH_(3)COOH, 0.2 mol` of `CH_(3)COONa`,and `0.05 mol` of `NaOH` in `1L. (pK_(a) of CH_(3)COOH = 4.74)` is:

Options

A. `5.44`
B. `5.20`
C. `5.04`
D. `4.74`

Explanation

To find the pH of the solution containing 0.1 mol of acetic acid (CH₃COOH), 0.2 mol of sodium acetate (CH₃COONa), and 0.05 mol of sodium hydroxide (NaOH) in 1 L, we can follow these steps:

### Step 1: Identify the components
We have:
- Acetic acid (CH₃COOH) - weak acid
- Sodium acetate (CH₃COONa) - salt of the weak acid, which will act as a base
- Sodium hydroxide (NaOH) - strong base

### Step 2: Determine the reaction
When NaOH is added to the solution, it will react with acetic acid:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 3: Calculate the initial moles
- Initial moles of CH₃COOH = 0.1 mol
- Initial moles of CH₃COONa = 0.2 mol
- Initial moles of NaOH = 0.05 mol

### Step 4: Determine the final moles after reaction
Since NaOH will react with CH₃COOH, we will subtract the moles of NaOH from the moles of CH₃COOH:
- Moles of CH₃COOH remaining = 0.1 mol - 0.05 mol = 0.05 mol
- Moles of CH₃COONa formed = Initial moles of CH₃COONa + moles from reaction with NaOH = 0.2 mol + 0.05 mol = 0.25 mol

### Step 5: Calculate the concentrations
Since the total volume of the solution is 1 L:
- Concentration of CH₃COOH = 0.05 mol / 1 L = 0.05 M
- Concentration of CH₃COONa = 0.25 mol / 1 L = 0.25 M

### Step 6: Use the Henderson-Hasselbalch equation
The pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Where:
- pKₐ of CH₃COOH = 4.74
- Base = CH₃COONa = 0.25 M
- Acid = CH₃COOH = 0.05 M

### Step 7: Substitute the values into the equation
\[ \text{pH} = 4.74 + \log\left(\frac{0.25}{0.05}\right) \]

### Step 8: Calculate the log term
\[ \frac{0.25}{0.05} = 5 \]
Thus,
\[ \log(5) \approx 0.7 \]

### Step 9: Final calculation of pH
\[ \text{pH} = 4.74 + 0.7 = 5.44 \]

### Final Answer
The pH of the solution is **5.44**.

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