Calculate the $pH$ of the following solutions:
$(a)$ $0.1 \ M \ HCl$
$(b)$ $0.1 \ M \ H_2SO_4$
$(c)$ $0.1 \ M \ HNO_3$
$(d)$ $0.1 \ M \ NaOH$
$(e)$ $0.1 \ M \ KOH$
$(f)$ $0.1 \ M \ Ba(OH)_2$

(A) For $HCl$ (strong acid),$[H^+] = 0.1 \ M = 10^{-1} \ M$. $pH = -\log[H^+] = -\log(10^{-1}) = 1$.
$(b)$ For $H_2SO_4$ (strong acid),$[H^+] = 2 \times 0.1 \ M = 0.2 \ M$. $pH = -\log(0.2) = -(\log 2 - 1) = 1 - 0.3010 = 0.6990$.
$(c)$ For $HNO_3$ (strong acid),$[H^+] = 0.1 \ M = 10^{-1} \ M$. $pH = -\log(10^{-1}) = 1$.
$(d)$ For $NaOH$ (strong base),$[OH^-] = 0.1 \ M = 10^{-1} \ M$. $pOH = -\log(10^{-1}) = 1$. $pH = 14 - pOH = 14 - 1 = 13$.
$(e)$ For $KOH$ (strong base),$[OH^-] = 0.1 \ M = 10^{-1} \ M$. $pOH = -\log(10^{-1}) = 1$. $pH = 14 - pOH = 14 - 1 = 13$.
$(f)$ For $Ba(OH)_2$ (strong base),$[OH^-] = 2 \times 0.1 \ M = 0.2 \ M$. $pOH = -\log(0.2) = 0.6990$. $pH = 14 - 0.6990 = 13.3010$.