Question

ính pH của dung dịch sau HNO3 0,05M; KOH 0,2M; H2SO4 0,1M.

Answer 1

\(ddHNO_3\\ \left[H^+\right]=\left[HNO_3\right]=0,05\left(M\right)\\ \Rightarrow pH=-log\left[0,05\right]\approx1,301\\ ddKOH\\ \left[OH^-\right]=\left[KOH\right]=14+log\left[0,2\right]=13,30103\\ ddH_2SO_4\\ \left[H^+\right]=2.0,1=0,2\left(M\right)\\ \Rightarrow pH=-log\left[0,2\right]\approx0,699\)