Câu 5: \(n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right);V_{dd}=200\left(ml\right)=0,2\left(l\right)\)
`=>` \(C_{M\left(NaOH\right)}=\dfrac{0,02}{0,2}=0,1M\Rightarrow\left[OH^-\right]=0,1=10^{-1}\)
`=>` \(pOH=1\Rightarrow pH=14-1=13\)
Câu 6:
Đặt `V_{ddNaOH} = x (l)`
`=>` \(\left\{{}\begin{matrix}n_{HCl}=0,094x\left(mol\right)\\n_{NaOH}=0,2.0,2=0,04\left(mol\right)\end{matrix}\right.\)
Vì dd sau phản ứng có `pH = 2 < 7 =>` axit dư
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
0,04---->0,04
`=>` \(n_{H^+\left(dư\right)}=n_{HCl\left(dư\right)}=0,094x-0,04\left(mol\right)\)
Ta có: \(pH=2\Rightarrow\left[H^+\right]=0,01\)
`=>` \(\left[H^+_{dư}\right]=\dfrac{0,094x-0,04}{0,2+x}=0,01\Leftrightarrow x=0,5\left(l\right)\)
`=> V = 500 (ml)`
Câu 8:
Ta có: \(\left\{{}\begin{matrix}n_{Ba\left(OH\right)_2}=0,005.0,02=0,0001\left(mol\right)\\n_{H_2SO_4}=0,08.0,002=0,00016\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)
0,0001----->0,0001
`=>` \(n_{H^+\left(dư\right)}=2n_{H_2SO_4\left(dư\right)}=\left(0,00016-0,0001\right).2=0,00012\left(mol\right)\)
`=>` \(\left[H^+\right]=\dfrac{0,00012}{0,02+0,08}=0,0012\left(mol\right)\)
`=>` \(pH=-log\left(0,0012\right)=2,921\)
giảm dần của axit ??
