To a 100 mL of 0.1 M weak acid HA solution 22.5 mL of 0.2 M solution of NaOH are added.Now, what volume of 0.1 M NaOH solution be added into above solution, so that pH of resulting solution be 4.7:
[Given :`(K_b(A^(-))=5xx10^(-10)]`

Options

A. 5 mL
B. 20 mL
C. 10 mL
D. 15 mL

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the initial moles of weak acid (HA) and NaOH added.
- **Given:**
- Volume of HA solution = 100 mL
- Molarity of HA solution = 0.1 M
- Volume of NaOH solution = 22.5 mL
- Molarity of NaOH solution = 0.2 M

- **Calculations:**
- Moles of HA = Volume (L) × Molarity (M) = \( 0.1 \, \text{L} \times 0.1 \, \text{mol/L} = 0.01 \, \text{mol} \) or 10 mmol.
- Moles of NaOH = Volume (L) × Molarity (M) = \( 0.0225 \, \text{L} \times 0.2 \, \text{mol/L} = 0.0045 \, \text{mol} \) or 4.5 mmol.

### Step 2: Determine the remaining moles of HA and the moles of salt (NaA) after the reaction.
- **Reaction:**
\[ \text{HA} + \text{NaOH} \rightarrow \text{NaA} + \text{H}_2\text{O} \]

- **Calculations:**
- Moles of HA remaining = Initial moles of HA - Moles of NaOH = \( 10 \, \text{mmol} - 4.5 \, \text{mmol} = 5.5 \, \text{mmol} \)
- Moles of NaA formed = Moles of NaOH used = 4.5 mmol.

### Step 3: Calculate the pKa from the given Kb.
- **Given:**
- \( K_b = 5 \times 10^{-10} \)

- **Using the relation:**
\[ K_a \times K_b = K_w \]
\[ K_a = \frac{K_w}{K_b} \]
Where \( K_w = 1 \times 10^{-14} \).

- **Calculations:**
\[ K_a = \frac{1 \times 10^{-14}}{5 \times 10^{-10}} = 2 \times 10^{-5} \]
\[ pK_a = -\log(K_a) = -\log(2 \times 10^{-5}) \approx 4.7 \]

### Step 4: Set up the equation for the buffer solution.
- **Using the Henderson-Hasselbalch equation:**
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

- **Substituting values:**
\[ 4.7 = pK_a + \log\left(\frac{4.5 + 0.1V}{5.5 - 0.1V}\right) \]

### Step 5: Solve for V.
- **Rearranging the equation:**
\[ 0 = \log\left(\frac{4.5 + 0.1V}{5.5 - 0.1V}\right) \]
\[ 1 = \frac{4.5 + 0.1V}{5.5 - 0.1V} \]

- **Cross-multiplying:**
\[ 5.5 - 0.1V = 4.5 + 0.1V \]
\[ 5.5 - 4.5 = 0.1V + 0.1V \]
\[ 1 = 0.2V \]
\[ V = \frac{1}{0.2} = 5 \, \text{mL} \]

### Final Answer:
The volume of 0.1 M NaOH solution that should be added is **5 mL**.

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