`10^(-2)` mole of `NaOH` was added to `10 litres` of water. The `pH` will change by

Options

A. 4
B. 3
C. 11
D. 7

Explanation

To solve the problem step-by-step, we will determine the change in pH after adding `10^(-2)` moles of `NaOH` to `10 liters` of water.

### Step 1: Determine the initial pH of water
- Pure water has a neutral pH of 7.

### Step 2: Calculate the molarity of `NaOH`
- Molarity (M) is calculated using the formula:
\[ \text{Molarity} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}} \]
- Given:
- Number of moles of `NaOH` = `10^(-2)` moles
- Volume of solution = `10 liters`

- Substituting the values:
\[ \text{Molarity of NaOH} = \frac{10^{-2}}{10} = 10^{-3} \, \text{M} \]

### Step 3: Determine the concentration of hydroxide ions (`OH-`)
- Since `NaOH` is a strong base, it fully dissociates in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
- Therefore, the concentration of `OH-` ions will also be `10^{-3} M`.

### Step 4: Calculate the pOH of the solution
- pOH is calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
- Substituting the concentration of `OH-`:
\[ \text{pOH} = -\log(10^{-3}) = 3 \]

### Step 5: Calculate the pH of the solution
- The relationship between pH and pOH is given by:
\[ \text{pH} + \text{pOH} = 14 \]
- Substituting the value of pOH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \]

### Step 6: Determine the change in pH
- The change in pH is calculated by subtracting the initial pH from the final pH:
\[ \text{Change in pH} = \text{Final pH} - \text{Initial pH} = 11 - 7 = 4 \]

### Final Answer
The change in pH after adding `10^(-2)` moles of `NaOH` to `10 liters` of water is **4**.

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