What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCI are mixed ?

Options

A. `7.0`
B. `1.04`
C. `12.65`
D. `2.0`

Answer

To find the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed, we can follow these steps:

### Step 1: Calculate the number of moles of NaOH and HCl
Assume we take 1 L (or any equal volume, V) of each solution for simplicity.

- Moles of NaOH = Concentration × Volume = 0.1 M × 1 L = 0.1 moles
- Moles of HCl = Concentration × Volume = 0.01 M × 1 L = 0.01 moles

### Step 2: Determine the reaction between NaOH and HCl
NaOH and HCl react in a 1:1 ratio to form NaCl and water:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 3: Calculate the remaining moles after the reaction
- Moles of NaOH remaining = Initial moles of NaOH - Moles of HCl = 0.1 moles - 0.01 moles = 0.09 moles
- Moles of HCl remaining = 0 (since all HCl is consumed)

### Step 4: Calculate the total volume of the mixture
Since we mixed equal volumes of both solutions:
- Total volume = Volume of NaOH + Volume of HCl = 1 L + 1 L = 2 L

### Step 5: Calculate the concentration of NaOH in the resulting solution
The concentration of NaOH in the resulting solution:
\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH remaining}}{\text{Total volume}} = \frac{0.09 \text{ moles}}{2 \text{ L}} = 0.045 \text{ M} \]

### Step 6: Calculate the pOH of the solution
Since NaOH is a strong base, the concentration of OH⁻ ions is equal to the concentration of NaOH:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log[0.045] \]

Calculating this gives:
\[ \text{pOH} \approx 1.35 \]

### Step 7: Calculate the pH of the solution
Using the relationship between pH and pOH:
\[ \text{pH} = 14 - \text{pOH} \]
\[ \text{pH} = 14 - 1.35 = 12.65 \]

### Final Answer:
The pH of the resulting solution is approximately **12.65**.

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