Calculate the pH of the solution prepared by dissolving 0.3 g of NaOH in water to give 200 mL of solution.

Answer

To calculate the pH of the solution prepared by dissolving 0.3 g of NaOH in water to give 200 mL of solution, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of NaOH (Sodium Hydroxide) is calculated as follows:
- Sodium (Na) = 23 g/mol
- Oxygen (O) = 16 g/mol
- Hydrogen (H) = 1 g/mol

Thus, the molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

Now, substituting the values:
\[ \text{Number of moles} = \frac{0.3 \text{ g}}{40 \text{ g/mol}} = 0.0075 \text{ moles} \]

### Step 2: Calculate the concentration of NaOH in the solution
Concentration (C) is given by the formula:
\[ C = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
The volume of the solution is 200 mL, which is equivalent to 0.2 L. Therefore:
\[ C = \frac{0.0075 \text{ moles}}{0.2 \text{ L}} = 0.0375 \text{ M} \]

### Step 3: Determine the concentration of OH⁻ ions
Since NaOH is a strong base, it dissociates completely in solution:
\[ \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^- \]
Thus, the concentration of OH⁻ ions is also 0.0375 M.

### Step 4: Calculate the pOH of the solution
The pOH is calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration of OH⁻:
\[ \text{pOH} = -\log(0.0375) \approx 1.426 \]

### Step 5: Calculate the pH of the solution
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
We can find the pH:
\[ \text{pH} = 14 - \text{pOH} = 14 - 1.426 \approx 12.574 \]

### Final Answer:
The pH of the solution is approximately **12.57**.

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