`f.`
`n_{H_2SO_4}=(4,9)/98=0,05(mol)`
`C_{M_{H_2SO_4}}=(0,05)/1=0,05(M)`
`H_2SO_4->2H^++SO_4^{2-}`
`[H^+]=2C_{M_{H_2SO_4}}=0,1(M)`
`pH=-log[H^+]=-log(0,1)=1`
`g.`
`n_{H_2SO_4}=0,1.0,01=0,001(mol)`
`n_{HCl}=0,1.0,05=0,005(mol)`
`V_{dd}=0,1+0,1=0,2(l)`
`C_{M_{H_2SO_4}}=(0,001)/(0,2)=0,005(M)`
`C_{M_{HCl}}=(0,005)/(0,2)=0,025(M)`
`H_2SO_4->2H^++SO_4^{2-}`
`HCl->H^++Cl^-`
`∑[H^+]=2C_{M_{H_2SO_4}}+C_{M_{HCl}}=0,035(M)`
`pH=-log[H^+]=-log(0,035)=1,456`
`h.`
`n_{KOH}=0,1.0,1=0,01(mol)`
`n_{Ba(OH)_2}=0,2.0,05=0,01(mol)`
`V_{dd}=0,1+0,2=0,3(l)`
`C_{M_{KOH}}=C_{M_{Ba(OH)_2}}=(0,01)/(0,3)=1/30(M)`
`KOH->K^++OH^-`
`Ba(OH)_2->Ba^{2+}+2OH^-`
`∑[OH^-]=C_{M_{KOH}}+2C_{M_{Ba(OH)_2}}=0,1(M)`
`pOH=-log[OH^-]=-log(0,1)=1`
`pH=14-pOH=14-1=13`
