What will be the `pH` of a solution formed by mixing `40 ml` of `0.10 M HCl` with `10 ml` of `0.45 M NaOH`?

Options

A. 12
B. 10
C. 8
D. 6

Answer

To find the pH of the solution formed by mixing 40 ml of 0.10 M HCl with 10 ml of 0.45 M NaOH, we can follow these steps:

### Step 1: Calculate the number of moles of HCl and NaOH

**For HCl:**
- Molarity (M) = 0.10 M
- Volume (V) = 40 ml = 0.040 L

Using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of HCl} = 0.10 \, \text{mol/L} \times 0.040 \, \text{L} = 0.004 \, \text{moles} \]

**For NaOH:**
- Molarity (M) = 0.45 M
- Volume (V) = 10 ml = 0.010 L

Using the same formula:
\[ \text{Moles of NaOH} = 0.45 \, \text{mol/L} \times 0.010 \, \text{L} = 0.0045 \, \text{moles} \]

### Step 2: Determine the limiting reactant and the remaining moles after the reaction

The reaction between HCl and NaOH is:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the calculations:
- Moles of HCl = 0.004 moles
- Moles of NaOH = 0.0045 moles

Since HCl is the limiting reactant (it has fewer moles), it will be completely consumed:
- Moles of NaOH remaining = 0.0045 - 0.004 = 0.0005 moles

### Step 3: Calculate the concentration of OH⁻ ions in the final solution

The total volume of the solution after mixing:
\[ \text{Total Volume} = 40 \, \text{ml} + 10 \, \text{ml} = 50 \, \text{ml} = 0.050 \, \text{L} \]

Now, calculate the concentration of OH⁻ ions:
\[ \text{Concentration of OH}^- = \frac{\text{Moles of OH}^-}{\text{Total Volume}} = \frac{0.0005 \, \text{moles}}{0.050 \, \text{L}} = 0.01 \, \text{M} \]

### Step 4: Calculate the pOH and then the pH

Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] = -\log[0.01] = 2 \]

Now, use the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
\[ \text{pH} = 14 - \text{pOH} = 14 - 2 = 12 \]

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

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