A solution is $0\mathrm{.}0300M$ in $N{H}_{4}Cl$ and $0\mathrm{.}0190M$ in $N{H}_3(K_a(N{H}_4^{+})=5\mathrm{.}70{10}^{-10}).$ Calculate

its $O{H}^{-}$concentration and its $pH$

a$.$ neglecting activities.

$O{H}^{-}$concentration $=$

M

$pH=$

b$.$ taking activities into account and $\{$:${}_{H_3{O}^{+}}=0\mathrm{.}9).$

$O{H}^{-}$concentration $=$

M

$pH=$

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