\(n_{KOH}=0,2.0,3=0,06mol\)
\(n_{H_2SO_4}=0,2.0,05=0,01mol\)
2KOH+H2SO4\(\rightarrow\)K2SO4+2H2O
\(\dfrac{0,06}{2}=0,03>\dfrac{0,01}{1}=0,01\)
KOH dư, H2SO4 hết
\(n_{KOH}\left(pu\right)=2n_{H_2SO_4}=0,02mol\)
\(n_{KOH\left(dư\right)}=0,06-0,02=0,04mol\)
Vdd=200+200=400ml=0,4 lít
\(C_{M_{KOH}}=\dfrac{n}{v}=\dfrac{0,04}{0,4}=0,1M\)
pH=14+lg[OH-]=14+lg0,1=13