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. `12.0`
B. `10.0`
C. `8.0`
D. `6.0`

Explanation

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 millimoles of HCl and NaOH.
- **HCl:**
- Volume = 40 mL
- Concentration = 0.10 M
- Millimoles of HCl = Volume (mL) × Concentration (M) = 40 mL × 0.10 M = 4 mmoles

- **NaOH:**
- Volume = 10 mL
- Concentration = 0.45 M
- Millimoles of NaOH = Volume (mL) × Concentration (M) = 10 mL × 0.45 M = 4.5 mmoles

### Step 2: Determine the limiting reactant and the reaction outcome.
The neutralization reaction between HCl and NaOH can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

From the stoichiometry of the reaction:
- 1 mole of HCl reacts with 1 mole of NaOH.
- We have 4 mmoles of HCl and 4.5 mmoles of NaOH.

Since HCl is the limiting reactant, it will completely react with 4 mmoles of NaOH, leaving:
- Remaining NaOH = 4.5 mmoles - 4 mmoles = 0.5 mmoles

### Step 3: Calculate the total volume of the solution after mixing.
- Total volume = Volume of HCl + Volume of NaOH = 40 mL + 10 mL = 50 mL

### Step 4: Calculate the concentration of the remaining NaOH.
- Concentration of NaOH after the reaction = Remaining millimoles / Total volume (in L)
- Remaining millimoles of NaOH = 0.5 mmoles
- Total volume = 50 mL = 0.050 L
- Concentration of NaOH = 0.5 mmoles / 0.050 L = 10 mM = 0.010 M

### Step 5: Calculate the pOH of the solution.
- pOH = -log[OH⁻]
- Since NaOH fully dissociates in solution, [OH⁻] = 0.010 M
- pOH = -log(0.010) = 2

### Step 6: Calculate the pH of the solution.
- pH + pOH = 14
- pH = 14 - pOH = 14 - 2 = 12

### Final Answer:
The pH of the solution formed by mixing 40 mL of 0.10 M HCl with 10 mL of 0.45 M NaOH is **12**.

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