Calculate `pH` of a solution whose `100ml` contains `0.2g NaOH` dissolved in it.

Options

A. `10.699`
B. `11.699`
C. `12.699`
D. `13.699`

Explanation

To calculate the pH of a solution containing 0.2 g of NaOH in 100 mL, follow these steps:

### Step 1: Calculate the number of moles of NaOH
First, we need to find the molar mass of NaOH.
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol

The molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

Now, we can calculate the number of moles of NaOH in 0.2 g:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} = \frac{0.2 \text{ g}}{40 \text{ g/mol}} = 0.005 \text{ moles} \]

### Step 2: Calculate the molarity of the NaOH solution
Molarity (M) is defined as the number of moles of solute per liter of solution. Since we have 100 mL of solution, we need to convert this to liters:
\[ 100 \text{ mL} = 0.1 \text{ L} \]

Now, we can calculate the molarity:
\[ \text{Molarity} = \frac{\text{number of moles}}{\text{volume of solution (L)}} = \frac{0.005 \text{ moles}}{0.1 \text{ L}} = 0.05 \text{ M} \]

### Step 3: Calculate the concentration of OH⁻ ions
Since NaOH is a strong base, it completely dissociates in solution:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]

Thus, the concentration of OH⁻ ions is equal to the molarity of NaOH:
\[ [\text{OH}^-] = 0.05 \text{ M} \]

### Step 4: Calculate pOH
To find pOH, we use the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration:
\[ \text{pOH} = -\log(0.05) = -\log(5 \times 10^{-2}) = -(\log(5) + \log(10^{-2})) = -(\log(5) - 2) \]
Using the approximation \(\log(5) \approx 0.7\):
\[ \text{pOH} \approx -(0.7 - 2) = 1.3 \]

### Step 5: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.3 = 12.7 \]

### Final Answer
The pH of the solution is approximately **12.7**.

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