\(n_{H_2SO_4}=0.2\cdot0.6=0.12\left(mol\right)\)
\(n_{KOH}=0.2\cdot1=0.2\left(mol\right)\)
\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)
Lập tỉ lệ :
\(\dfrac{n_{KOH}}{2}< \dfrac{n_{H_2sO_4}}{1}\) \(\Rightarrow H_2SO_4\) dư
\(V_{dd}=0.2+0.2=0.4\left(l\right)\)
\(n_{H_2SO_4\left(dư\right)}=0.2-\dfrac{0.12}{2}=0.14\left(mol\right)\)
\(\left[H^+\right]=2\cdot\dfrac{0.14}{0.4}=0.7\left(M\right)\)