Câu trả lời:
\(n_{ }_{NAOH}=0.2.1=0.2mol\)
\(n_{H_2sõ_4}=0.2.2=0.4mol\)
\(OH^-+H^+\rightarrow H_20\)
0.2 0.8
\(NAOH\rightarrow Na^++OH^-\)
0.2 0.2
\(H_2SO_4\rightarrow2H^++SO^{2-}_4\)
0.4 0.4
\(\left[Na^+\right]=\frac{0.2}{0.4}=0.5M\)
\(\left[SO^{2-}_4\right]=\frac{0.4}{0.4}=1M\)
\(\left[H^+_{dư}\right]=\frac{0,6}{0,4}=1.5M\)