1) Ta coi H2SO4 điện li mạnh hai nấc.
\(n_{H_2SO_4}=0,03\) mol; \(n_{HCl}=0,16\) mol
\(H_2SO_4\rightarrow2H^++SO_4^-\)
0,03 -----> 0,06 ---> 0,03
\(HCl\rightarrow H^++Cl^-\)
0,16 --> 0,16 --> 0,16
\(\Rightarrow n_{H^+}=2n_{H_2SO_4}+n_{HCl}=2.0,03+0,16=0,22\) mol
+ \(\left[H^+\right]=\dfrac{0,22}{1+4}=0,044\) mol/lít
\(\Rightarrow pH=-lg\left[H^+\right]=-lg0,044=1,36\)
+ \(\left[SO_4^-\right]=\dfrac{0,03}{1+4}=6.10^{-3}\) mol/lít
+ \(\left[Cl^-\right]=\dfrac{0,16}{1+4}=0,032\) mol/lít
2) + \(n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,12\) mol
\(\Rightarrow\left[OH^-\right]=\dfrac{0,12}{5}=0,024\) mol/lít
\(\Rightarrow pOH=-lg\left[OH^-\right]=1,62\)
\(\Rightarrow pH=14-pOH=12,38\)
+ \(n_{Na^+}=n_{NaOH}=0,06\) mol
\(\Rightarrow\left[Na^+\right]=\dfrac{0,06}{5}=0,012\) mol/lít
+ \(n_{Ba^{2+}}=n_{Ba\left(OH\right)_2}=0,03\) mol
\(\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,03}{5}=0,006\) mol/lít