Question

The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is $x \times 10^{-1}$. The value of $x$ is ________. (Nearest integer) [log 2.5 = 0.3979]

Hint:

For mixtures of strong acids and bases, always work with millimoles (mL $\times$ M) to avoid dealing with small fractions until the final step.

Answer 1

Correct Answer: 6021

Step 1: Understanding the Concept:
When a strong acid and a strong base react, neutralization occurs. The pH of the resulting solution depends on the concentration of the excess reactant (either $\text{H}^+$ or $\text{OH}^-$) in the final total volume.
Step 2: Detailed Explanation:
1. Calculate millimoles (mmol) of reactants:
- $\text{mmol of H}^+ = 50 \text{ mL} \times 1 \text{ M} = 50 \text{ mmol}$
- $\text{mmol of OH}^- = 30 \text{ mL} \times 1 \text{ M} = 30 \text{ mmol}$
2. Acid is in excess. Excess $\text{H}^+ = 50 - 30 = 20 \text{ mmol}$.
3. Total volume of solution $= 50 + 30 = 80 \text{ mL}$.
4. Final concentration of $\text{H}^+$:
\[ [\text{H}^+] = \frac{20 \text{ mmol}}{80 \text{ mL}} = 0.25 \text{ M} = 2.5 \times 10^{-1} \text{ M} \]
5. Calculate pH:
\[ \text{pH} = -\log [\text{H}^+] = -\log (2.5 \times 10^{-1}) = 1 - \log 2.5 \]
\[ \text{pH} = 1 - 0.3979 = 0.6021 \]
6. Comparing with $x \times 10^{-1}$:
\[ 0.6021 = 6.021 \times 10^{-1} \implies x \approx 6 \]
Step 3: Final Answer:
The value of $x$ is 6.