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 of finding the pH of a diluted NaOH solution, we can follow these steps:

### Step 1: Understand the initial conditions
We have a 100 mL solution of NaOH with a concentration of 0.01 M. We need to dilute this solution to a total volume of 1 liter (1000 mL).

### Step 2: Use the dilution formula
The dilution formula is given by:
\[ M_1 V_1 = M_2 V_2 \]
Where:
- \( M_1 \) = initial molarity (0.01 M)
- \( V_1 \) = initial volume (100 mL)
- \( M_2 \) = final molarity (unknown)
- \( V_2 \) = final volume (1000 mL)

### Step 3: Substitute the known values
Convert volumes to liters for consistency:
- \( V_1 = 100 \, \text{mL} = 0.1 \, \text{L} \)
- \( V_2 = 1000 \, \text{mL} = 1 \, \text{L} \)

Now substituting the values into the dilution formula:
\[ 0.01 \, \text{M} \times 0.1 \, \text{L} = M_2 \times 1 \, \text{L} \]

### Step 4: Solve for \( M_2 \)
\[ M_2 = \frac{0.01 \times 0.1}{1} = 0.001 \, \text{M} \]

### Step 5: Determine the concentration of hydroxide ions
Since NaOH is a strong base, it completely dissociates in solution:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Thus, the concentration of hydroxide ions \([OH^-]\) is equal to the molarity of NaOH:
\[ [OH^-] = 0.001 \, \text{M} \]

### Step 6: Calculate the pOH
Using the formula:
\[ \text{pOH} = -\log[OH^-] \]
Substituting the value:
\[ \text{pOH} = -\log(0.001) = 3 \]

### Step 7: Calculate the pH
Using the relationship:
\[ \text{pH} + \text{pOH} = 14 \]
We can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \]

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

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