`4g` of `NaOH` was dissolved in one litre of a solution containing one mole of acetic acid and one mole of sodium acetete. Find the pH of the resulting solution (`K_(a)` of acetic acid is `1.8xx10^(-3)`)

Options

A. `4.70`
B. `4.83`
C. `5.92`
D. `6.87`

Explanation

To find the pH of the resulting solution after dissolving 4g of NaOH in a solution containing one mole of acetic acid and one mole of sodium acetate, we can follow these steps:

### Step 1: Calculate the moles of NaOH
First, we need to calculate the number of moles of NaOH present in 4g.

- Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol
- Moles of NaOH = mass (g) / molar mass (g/mol) = 4g / 40 g/mol = 0.1 moles

### Step 2: Determine the reaction with acetic acid
NaOH is a strong base and will react with acetic acid (CH₃COOH), which is a weak acid. The reaction can be represented as follows:

\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

Since we have 1 mole of acetic acid and 0.1 moles of NaOH, the NaOH will react with 0.1 moles of acetic acid.

### Step 3: Calculate the remaining moles of acetic acid and sodium acetate
After the reaction, the moles of acetic acid and sodium acetate will change as follows:

- Moles of acetic acid remaining = 1 mole - 0.1 moles = 0.9 moles
- Moles of sodium acetate formed = 1 mole + 0.1 moles = 1.1 moles

### Step 4: Use the Henderson-Hasselbalch equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]

Where:
- \(\text{pK}_a = -\log(K_a)\)
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate)
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid)

### Step 5: Calculate pK_a
Given \( K_a = 1.8 \times 10^{-5} \):

\[ \text{pK}_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 6: Calculate concentrations
Since the total volume of the solution is 1 L, the concentrations are:

- \([\text{A}^-] = [\text{CH}_3\text{COO}^-] = 1.1 \, \text{mol/L}\)
- \([\text{HA}] = [\text{CH}_3\text{COOH}] = 0.9 \, \text{mol/L}\)

### Step 7: Substitute values into the Henderson-Hasselbalch equation
Now, substituting the values into the equation:

\[ \text{pH} = 4.74 + \log \left( \frac{1.1}{0.9} \right) \]

Calculating the logarithm:

\[ \log \left( \frac{1.1}{0.9} \right) \approx \log(1.222) \approx 0.087 \]

So,

\[ \text{pH} = 4.74 + 0.087 \approx 4.83 \]

### Final Answer
The pH of the resulting solution is approximately **4.83**.

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