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

Options

A. 4
B. 7
C. 3
D. 8

Explanation

To solve the problem of how much the pH will change when `10^(-3)` moles of `NaOH` is added to `10 liters` of water, we will follow these steps:

### Step 1: Calculate the concentration of `NaOH`
We know that concentration (molarity) is calculated as:
\[ \text{Concentration (M)} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
In this case, we have:
- Number of moles of `NaOH` = \(10^{-3}\) moles
- Volume of water = \(10\) liters

Thus, the concentration of `NaOH` will be:
\[ \text{Concentration} = \frac{10^{-3}}{10} = 10^{-4} \text{ M} \]

### Step 2: Determine the concentration of `OH⁻` ions
Since `NaOH` is a strong base, it completely dissociates in water:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
This means that the concentration of `OH⁻` ions will also be \(10^{-4} \text{ M}\).

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

### Step 4: Calculate the pH
We know that:
\[ \text{pH} + \text{pOH} = 14 \]
So, we can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 4 = 10 \]

### Step 5: Determine the change in pH
The initial pH of pure water is \(7\). Therefore, the change in pH (ΔpH) when `NaOH` is added is:
\[ \Delta \text{pH} = \text{Final pH} - \text{Initial pH} = 10 - 7 = 3 \]

### Conclusion
Thus, the pH will change by **3**.

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