Question

At 27°C, 100 mL of 0.4 M HCl is mixed with 100 mL of 0.5 M NaOH. To the resultant solution, 800 mL of distilled water is added. What is the pH of the final solution?

• A. \( 12 \)
• B. \( 2 \)
• C. \( 1.3 \)
• D. \( 1.0 \)

Hint:

When mixing a strong acid and a strong base, first determine the limiting reactant. The excess species will dictate the final pH. If excess NaOH remains, use its concentration to find pOH, then convert to pH using \( pH + pOH = 14 \).

Answer 1

The Correct Option is A

To determine the pH of the final solution, follow these steps:

1. Identify the initial moles of HCl and NaOH:
- Moles of HCl = \(0.4 \, \text{mol/L} \times 0.1 \, \text{L} = 0.04 \, \text{mol}\)
- Moles of NaOH = \(0.5 \, \text{mol/L} \times 0.1 \, \text{L} = 0.05 \, \text{mol}\)

2. Determine the reaction between HCl and NaOH:
- HCl reacts with NaOH in a 1:1 ratio: \( \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)
- HCl is the limiting reactant, so 0.04 mol of NaOH will react, leaving \(0.05 - 0.04 = 0.01 \, \text{mol of NaOH}\) unreacted.

3. Volume of the resultant solution:
- Total volume after mixing HCl and NaOH = \(0.1 \, \text{L} + 0.1 \, \text{L} = 0.2 \, \text{L}\)
- After adding 800 mL of distilled water, the total volume = \(0.2 \, \text{L} + 0.8 \, \text{L} = 1 \, \text{L}\)

4. Concentration of excess NaOH in the final solution:
- Since 0.01 mol of NaOH remains in 1 L, the concentration of NaOH = \(0.01 \, \text{mol/L}\)

5. Calculate the pH of the final solution:
- NaOH is a strong base, so \(\text{pOH} = -\log[0.01] = 2\).
- Using the relationship \(\text{pH} + \text{pOH} = 14\), we find \(\text{pH} = 14 - 2 = 12\).

Thus, the pH of the final solution is 12.

Answer 2

The given reaction involves mixing HCl and NaOH solutions. First, let's calculate the moles of each:

Moles of HCl = Volume × Concentration = 0.1 L × 0.4 M = 0.04 moles 

Moles of NaOH = Volume × Concentration = 0.1 L × 0.5 M = 0.05 moles

Since NaOH is in excess, it will neutralize all the HCl, and the remaining moles of NaOH are:

Excess moles of NaOH = 0.05 − 0.04 = 0.01 moles

Now, the total volume of the solution after adding 800 mL of distilled water is:

Total volume = 100 mL (HCl) + 100 mL (NaOH) + 800 mL (water) = 1 L

Concentration of OH⁻ ions in the final solution:

[OH⁻] = 0.01 moles / 1 L = 0.01 M

Calculate pOH:
pOH = −log[OH⁻] = −log(0.01) = 2

Calculate pH:
pH = 14 − pOH = 14 − 2 = 12

Final Answer: The pH of the resulting solution is 12.