Question

Equal volumes of pH 3, 4, and 5 are mixed in a container. The concentration of \( {H}^+ \) in the mixture is (Assume there is no change in the volume during mixing):

• A. \( 1 \times 10^{-3} \, {M} \)
• B. \( 3.7 \times 10^{-4} \, {M} \)
• C. \( 1 \times 10^{-4} \, {M} \)
• D. \( 3.7 \times 10^{-5} \, {M} \)
• E. \( 3 \times 10^{-5} \, {M} \)

Hint:

When mixing equal volumes of solutions with different pH, the concentration of \( {H}^+ \) is the average of the individual concentrations.

Answer 1

The Correct Option is B

To calculate the concentration of \( {H}^+ \) ions in the mixture, we first determine the concentration of \( {H}^+ \) for each solution:
- pH 3: \( [{H}^+] = 10^{-3} \, {M} \)
- pH 4: \( [{H}^+] = 10^{-4} \, {M} \)
- pH 5: \( [{H}^+] = 10^{-5} \, {M} \)
Since equal volumes of each solution are mixed, the average concentration of \( {H}^+ \) ions is the arithmetic mean of the individual concentrations: \[ [{H}^+]_{{mixture}} = \frac{1}{3} \left( 10^{-3} + 10^{-4} + 10^{-5} \right) \] Calculating this: \[ [{H}^+]_{{mixture}} = \frac{1}{3} \left( 0.001 + 0.0001 + 0.00001 \right) = \frac{1}{3} \times 0.00111 = 3.7 \times 10^{-4} \, {M} \] Thus, the concentration of \( {H}^+ \) in the mixture is \( 3.7 \times 10^{-4} \, {M} \).