100 mL of 0.01 M solution of NaOH is diluted to 1 litre. The pH of resultant solution will be

Options

A. 3
B. 12
C. 11
D. 8

Answer

To solve the problem step by step, we will follow the dilution formula and calculate the pH of the resultant solution.

### Step 1: Calculate the number of moles of NaOH in the initial solution.
The number of moles (n) can be calculated using the formula:
\[ n = \text{Molarity} \times \text{Volume} \]

Given:
- Molarity (C) = 0.01 M
- Volume (V) = 100 mL = 0.1 L (since 1 L = 1000 mL)

Calculating the number of moles:
\[ n = 0.01 \, \text{mol/L} \times 0.1 \, \text{L} = 0.001 \, \text{mol} \]

### Step 2: Determine the final concentration after dilution.
When the solution is diluted to 1 L, the number of moles remains the same, but the volume changes.

Using the dilution formula:
\[ C_1V_1 = C_2V_2 \]
Where:
- \( C_1 \) = initial concentration (0.01 M)
- \( V_1 \) = initial volume (0.1 L)
- \( C_2 \) = final concentration (unknown)
- \( V_2 \) = final volume (1 L)

Rearranging the formula to find \( C_2 \):
\[ C_2 = \frac{C_1V_1}{V_2} \]
\[ C_2 = \frac{0.01 \, \text{mol/L} \times 0.1 \, \text{L}}{1 \, \text{L}} = 0.001 \, \text{mol/L} \]

### Step 3: Calculate the concentration of OH⁻ ions.
Since NaOH is a strong base, it completely dissociates in solution:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

Thus, the concentration of OH⁻ ions is equal to the concentration of NaOH:
\[ [\text{OH}^-] = 0.001 \, \text{mol/L} \]

### Step 4: Calculate pOH.
The pOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]

Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.001) = -\log(10^{-3}) = 3 \]

### Step 5: Calculate pH.
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]

Substituting the value of pOH:
\[ \text{pH} + 3 = 14 \]
\[ \text{pH} = 14 - 3 = 11 \]

### Final Answer:
The pH of the resultant solution after dilution is **11**.

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