Calculate the $pH$ of the following solutions:
$(a)$ $0.002 \ M \ HNO_3$
$(b)$ $0.06 \ M \ H_2SO_4$
$(a)$ $0.002 \ M \ HNO_3$
$(b)$ $0.06 \ M \ H_2SO_4$
(N/A) For $HNO_3$,which is a strong monoprotic acid,$[H^+] = [HNO_3] = 0.002 \ M$.
$pH = -\log[H^+] = -\log(2 \times 10^{-3}) = 3 - \log 2 = 3 - 0.3010 = 2.699$.
$(b)$ For $H_2SO_4$,which is a strong diprotic acid,$[H^+] = 2 \times [H_2SO_4] = 2 \times 0.06 = 0.12 \ M$.
$pH = -\log[H^+] = -\log(0.12) = -\log(1.2 \times 10^{-1}) = 1 - \log 1.2 = 1 - 0.0792 = 0.9208$.
$pH = -\log[H^+] = -\log(2 \times 10^{-3}) = 3 - \log 2 = 3 - 0.3010 = 2.699$.
$(b)$ For $H_2SO_4$,which is a strong diprotic acid,$[H^+] = 2 \times [H_2SO_4] = 2 \times 0.06 = 0.12 \ M$.
$pH = -\log[H^+] = -\log(0.12) = -\log(1.2 \times 10^{-1}) = 1 - \log 1.2 = 1 - 0.0792 = 0.9208$.
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