`a)H^[+] + OH^[-] -> H_2 O`
`0,03` `0,03` `(mol)`
`n_[OH^-]=(0,1+0,1.2).0,1=0,03(mol)`
`=>V_[dd HCl]=[0,03]/[0,3]=0,1(l)`
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`b)H^[+] + OH^[-] -> H_2 O`
`0,02` `0,02`
`n_[H^+]=0,1.0,1+0,05.2.0,1=0,02(mol)`
`=>V_[dd NaOH]=[0,02]/[0,1]=0,2(l)`
\(a,H^++OH^-\rightarrow H_2O\\ n_{OH^-}=n_{NaOH}+n_{Ba\left(OH\right)_2}.2=0,1.0,1+2.0,1.0,1=0,03\left(mol\right)\\ \Rightarrow n_{H^+}=n_{OH^-}=0,03\left(mol\right)=n_{HCl}\\ V_{ddHCl}=\dfrac{0,03}{0,3}=0,1\left(l\right)\\ b,H^++OH^-\rightarrow H_2O\\ n_{H^+}=n_{HNO_3}+2.n_{H_2SO_4}=0,1.0,1+2.0,05.0,1=0,02\left(mol\right)=n_{OH^-}=n_{NaOH}\\ V_{ddNaOH}=\dfrac{0,02}{0,1}=0,2\left(l\right)\)