Calculate the change in pH of water when 0.01 mole of NaOH are added in 10 litre water.

Answer

To solve the problem of calculating the change in pH of water when 0.01 moles of NaOH are added to 10 liters of water, we can follow these steps:

### Step 1: Calculate the concentration of NaOH
To find the concentration of NaOH in the solution, we use the formula:

\[ \text{Concentration (M)} = \frac{\text{Number of moles}}{\text{Volume (L)}} \]

Given:
- Number of moles of NaOH = 0.01 moles
- Volume of water = 10 liters

Substituting the values:

\[ \text{Concentration of NaOH} = \frac{0.01 \text{ moles}}{10 \text{ L}} = 0.001 \text{ M} \]

### Step 2: Calculate the concentration of hydroxide ions (OH⁻)
Since NaOH is a strong base, it completely dissociates in water:

\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

Thus, the concentration of hydroxide ions (OH⁻) will also be 0.001 M.

### Step 3: Calculate the pOH of the solution
To find the pOH of the solution, we use the formula:

\[ \text{pOH} = -\log[\text{OH}^-] \]

Substituting the concentration of OH⁻:

\[ \text{pOH} = -\log(0.001) = 3 \]

### Step 4: Calculate the pH of the solution
We can find the pH using the relationship between pH and pOH:

\[ \text{pH} + \text{pOH} = 14 \]

Substituting the value of pOH:

\[ \text{pH} = 14 - \text{pOH} = 14 - 3 = 11 \]

### Step 5: Calculate the change in pH
Initially, the pH of pure water is 7. After adding NaOH, the pH is 11. Therefore, the change in pH is:

\[ \text{Change in pH} = \text{Final pH} - \text{Initial pH} = 11 - 7 = 4 \]

### Final Answer
The change in pH of the water when 0.01 mole of NaOH is added to 10 liters of water is **4**.

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