0.023 g of sodium metal is reacted with 100`cm^(3)` of water. The pH of the resulting solution is

Options

A. 10
B. 11
C. 9
D. 12

Answer

To solve the problem of determining the pH of the solution formed when 0.023 g of sodium metal reacts with 100 cm³ of water, we can follow these steps:

### Step 1: Write the Reaction
When sodium (Na) reacts with water (H₂O), it produces sodium hydroxide (NaOH) and hydrogen gas (H₂):
\[ 2 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2 \]

### Step 2: Calculate the Moles of Sodium
To find the number of moles of sodium, use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of sodium (Na) is 23 g/mol. Thus:
\[ \text{Number of moles of Na} = \frac{0.023 \, \text{g}}{23 \, \text{g/mol}} = 0.001 \, \text{moles} \]

### Step 3: Determine Moles of Sodium Hydroxide Produced
From the reaction, we see that 1 mole of sodium produces 1 mole of sodium hydroxide (NaOH). Therefore:
\[ \text{Moles of NaOH} = \text{Moles of Na} = 0.001 \, \text{moles} \]

### Step 4: Calculate the Concentration of OH⁻ Ions
Since NaOH dissociates completely in water to give OH⁻ ions:
\[ \text{Moles of OH⁻} = \text{Moles of NaOH} = 0.001 \, \text{moles} \]
The volume of the solution is 100 cm³, which is equivalent to 0.1 L. Therefore, the concentration of OH⁻ ions is:
\[ \text{Concentration of OH⁻} = \frac{0.001 \, \text{moles}}{0.1 \, \text{L}} = 0.01 \, \text{M} \]

### Step 5: Calculate the pOH of the Solution
The pOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH⁻}] \]
Substituting the concentration:
\[ \text{pOH} = -\log(0.01) = 2 \]

### Step 6: Calculate the pH of the Solution
Using the relationship between pH and pOH at 25°C:
\[ \text{pH} + \text{pOH} = 14 \]
Thus:
\[ \text{pH} = 14 - \text{pOH} = 14 - 2 = 12 \]

### Final Answer
The pH of the resulting solution is **12**.

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