(B) Using the dilution formula $M_1V_1 = M_2V_2$: $0.1 \ M \times 25 \ mL = M_2 \times 500 \ mL$ $M_2 = \frac{0.1 \times 25}{500} = 0.005 \ M$ Since $

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(B) Using the dilution formula $M_1V_1 = M_2V_2$: $0.1 \ M \times 25 \ mL = M_2 \times 500 \ mL$ $M_2 = \frac{0.1 \times 25}{500} = 0.005 \ M$ Since $HCl$ is a strong acid,$[H^+] = [HCl] = 0.005 \ M = 5 \times 10^{-3} \ M$. $pH = -\log[H^+] = -\log(5 \times 10^{-3}) = 3 - \log 5 = 3 - 0.698 = 2.302$.