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

Options

A. (a)10.699
B. (b)11.699
C. (c)12.699
D. (d)13.699

Answer

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 number of moles of NaOH using the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of NaOH (sodium hydroxide) is approximately 40 g/mol.

Given:
- Mass of NaOH = 0.2 g

Calculating the number of moles:
\[ \text{Number of moles} = \frac{0.2 \, \text{g}}{40 \, \text{g/mol}} = 0.005 \, \text{moles} \]

### 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 volume to liters:
\[ \text{Volume in liters} = \frac{100 \, \text{mL}}{1000} = 0.1 \, \text{L} \]

Now, we can calculate the molarity:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} = \frac{0.005 \, \text{moles}}{0.1 \, \text{L}} = 0.05 \, \text{M} \]

### 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 the NaOH solution:
\[ [\text{OH}^-] = 0.05 \, \text{M} \]

### Step 4: Calculate the pOH of the solution
Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration:
\[ \text{pOH} = -\log(0.05) = -\log(5 \times 10^{-2}) = -(\log 5 + \log 10^{-2}) = -\log 5 + 2 \]
Using the approximate value \(\log 5 \approx 0.69\):
\[ \text{pOH} = 2 - 0.69 = 1.31 \]

### Step 5: Calculate the pH of the solution
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Substituting the pOH value:
\[ \text{pH} = 14 - 1.31 = 12.69 \]

### Final Answer
The pH of the solution is approximately **12.69**.

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