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

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

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, the remaining moles of NaOH after neutralizing HCl 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

The concentration of OH^- ions in the final solution is:

[OH^-] = moles of NaOH / total volume = 0.01 moles / 1 L = 0.01 M

To find the pH, we first calculate the pOH using the formula:

pOH = -log[OH^-] = -log(0.01) = 2

Finally, pH is related to pOH by the equation:

pH + pOH = 14

pH = 14 - pOH = 14 - 2 = 12