Question

Equal volumes of two solutions A and B of a strong acid having pH = 6.0 and pH = 4.0 respectively are mixed together to form a new solution. The pH of the new solution will be in the range:

• A. Between 5 and 6
• B. Between 6 and 7
• C. Between 4 and 5
• D. Between 3 and 4

Hint:

Key Points: For strong acids: pH = $-\log[\text{H^+]$ When mixing solutions: concentrations average (for equal volumes) The more acidic solution dominates the final pH pH calculation: $-\log(0.5 \times 10^{-4) = 4 - \log(0.5) \approx 4.3$

Answer 1

The Correct Option is C

For strong acids:

• Solution A (pH = 6.0): [H⁺] = 10^-6 M
• Solution B (pH = 4.0): [H⁺] = 10^-4 M

When equal volumes are mixed: \[ \text{[H}^+\text{]}_{\text{new}} = \frac{10^{-6} + 10^{-4}}{2} \approx \frac{10^{-4}}{2} = 0.5 \times 10^{-4} \text{ M} \] Calculating pH of the new solution: \[ \text{pH} = -\log(0.5 \times 10^{-4}) = 4 - \log(0.5) \approx 4.3 \]
Thus, the pH falls between 4 and 5.