0.05 mole of NaOH is added to 5 liters of water What will be the pH of the solution ?

Options

A. a.12
B. b.7
C. c. 2
D. d. 10

Explanation

To find the pH of a solution when 0.05 moles of NaOH is added to 5 liters of water, we can follow these steps:

### Step 1: Calculate the concentration of NaOH
To find the concentration (C) of NaOH in the solution, we use the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume in liters}} \]
Given:
- Number of moles of NaOH = 0.05 moles
- Volume of water = 5 liters

Substituting the values:
\[ C = \frac{0.05 \text{ moles}}{5 \text{ liters}} = 0.01 \text{ M} \]

### Step 2: Write the dissociation equation
When NaOH is dissolved in water, it dissociates completely into sodium ions (Na⁺) and hydroxide ions (OH⁻):
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Since NaOH dissociates completely, the concentration of hydroxide ions (OH⁻) will also be 0.01 M.

### Step 3: Calculate the concentration of hydrogen ions (H⁺)
We can use the ion product of water (Kw) to find the concentration of hydrogen ions:
\[ K_w = [H^+][OH^-] \]
At 25°C, \(K_w = 1 \times 10^{-14}\).

We know the concentration of hydroxide ions:
\[ [OH^-] = 0.01 \text{ M} = 1 \times 10^{-2} \text{ M} \]

Now, we can rearrange the equation to find the concentration of hydrogen ions:
\[ [H^+] = \frac{K_w}{[OH^-]} = \frac{1 \times 10^{-14}}{1 \times 10^{-2}} = 1 \times 10^{-12} \text{ M} \]

### Step 4: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of [H⁺]:
\[ \text{pH} = -\log(1 \times 10^{-12}) = 12 \]

### Conclusion
The pH of the solution after adding 0.05 moles of NaOH to 5 liters of water is **12**.

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