$a)n_{NH_4Cl}=\dfrac{1}{53,5}=18,7.10^{-3}(mol);n_{Ba(OH)_2}=\dfrac{1}{315}=3,17.10^{-3}(mol)$

$NH_4^{+}+OH^{-}\rightarrow NH_3+H_2O$

$6,34.10^{-3} \rightarrow 6,34.10^{-3} \rightarrow 6,34.10^{-3}$
$\rightarrow n_{NH_4^+ dư}=18,7.10^{-3}-6,34.10^{-3}=12,36.10^{-3}(mol)$

$NH_4^{+}\leftrightharpoons NH_3+H^+$
$K_a=\dfrac{[NH_3][H^+]}{[NH_4^+]} $$\rightarrow [H^+]=\dfrac{10^{-9,24}.12,36.10^{-2}}{6,34.10^{-2}}=1,12.10^{-9}M$

$\rightarrow pH=8,95$

$b) [Ba^{2+}]=0,0317M;[NH_4^+]=0,1236M;[H^+]=1,12.10^{-9}M;[Cl^-]=0,187M$

$[OH^-]=\dfrac{10^{-14}}{1,12.10^{-9}}=8,93.10^{-6}M$

$c) NH_3+H^{+}\rightarrow NH_4^+$
$\rightarrow n_{H^+ dư}=0,01.1-6,34.10^{-3}=0,0366(mol)$

bỏ qua sự phân ly $NH_4^+$ do k đáng kể $\rightarrow [H^+]=\dfrac{0,0366}{0,11}=0,0333M \rightarrow pH=1,48$