Ok, để thử coi chứ tui ngu hóa thấy mồ :(

a/ \(n_{NaOH}=0,2.0,1=0,02\left(mol\right)\)

\(NaOH\rightarrow Na^++OH^-\)

\(n_{Na^+}=n_{OH^-}=0,02\left(mol\right)\)

\(\Rightarrow C_{MNa^+}=\frac{0,02}{0,4+0,1}=0,04\left(mol/l\right)\)

\(n_{Ba\left(OH\right)_2}=0,3.0,4=0,12\left(mol\right)\)

\(Ba\left(OH\right)_2=Ba^{2+}+2OH^-\)

\(\Rightarrow n_{OH^-}=0,24\left(mol\right);n_{Ba^{2+}}=0,12\left(mol\right)\)

\(\Rightarrow C_{MBa^{2+}}=\frac{0,12}{0,5}=0,24\left(mol/l\right)\)

\(n_{OH^-}=0,02+0,24=0,26\left(mol\right)\)

\(\Rightarrow C_{MOH^-}=\frac{0,26}{0,5}=0,52\left(mol/l\right)\)

b/ \(n_{HCl}=0,2V\left(mol\right)\)

\(\Rightarrow n_{H^+}=n_{Cl^-}=0,2V\)

\(\Rightarrow C_{MCl^-}=\frac{0,2V}{2V}=0,1\left(mol/l\right)\)

\(n_{H_2SO_4}=0,3V\left(mol\right)=\frac{n_{H^+}}{2}=n_{SO_4^{2-}}\)

\(\Rightarrow C_{MSO_4^{2-}}=\frac{0,3V}{2V}=0,15\left(mol/l\right)\)

\(n_{H^+}=0,2V+0,6V=0,8V\left(mol\right)\)

\(\Rightarrow C_{MH^+}=\frac{0,8V}{2V}=0,4\left(mol/l\right)\)

Bác nào hảo tâm giúp em mấy câu còn lại chớ đến đây thì em chịu chết òi :(

\(n_{H^+}=0,2.0,3=0,06\left(mol\right)\)

\(n_{OH^-}=0,5.2.0,001V=0,001V\left(mol\right)\)

\(\Rightarrow n_{OH^-dư}=0,001V-0,06\left(mol\right)\)

\(\left[OH^-\right]=\dfrac{0,001V-0,06}{0,001V+0,3}=0,28\)

\(\Rightarrow V=200\left(ml\right)\)

\(n_{OH^-}=0,5.0,2+0,2.2.0,3=0,22\left(mol\right)\Rightarrow\left[OH^-\right]=\dfrac{0,22}{0,5}=0,44M\)

\(n_{Na^+}=0,5.0,2=0,1\left(mol\right)\Rightarrow\left[Na^+\right]=\dfrac{0,1}{0,5}=0,2M\)

\(n_{Ba^{2+}}=0,2.0,3=0,06\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{0,06}{0,5}=0,12M\)

\(b.n_{NaOH\left(tổng\right)}=0,4.0,5+\dfrac{100.1,33.20\%}{40}=0,865\left(mol\right)\\ \left[Na^+\right]=\left[OH^-\right]=\left[NaOH\left(sau\right)\right]=\dfrac{0,865}{0,4+0,1}=1,73\left(M\right)\\ c.n_{HCl}=0,05.0,12=0,006\left(mol\right)\\ n_{HNO_3}=0,15.0,1=0,015\left(mol\right)\\ \left[H^+\right]=\dfrac{0,006+0,015}{0,05+0,15}=0,105\left(M\right)\\ \left[NO^-_3\right]=\dfrac{0,015}{0,05+0,15}=0,075\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,006}{0,05+0,15}=0,03\left(M\right)\)

\(d.n_{H_2SO_4}=0,4.0,05=0,02\left(mol\right)\\ n_{HCl}=0,35.0,2=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,05+0,35}=0,05\left(M\right)\\ \left[Cl^-\right]=\dfrac{0,07}{0,05+0,35}=0,175\left(M\right)\\ f.n_{KOH}=\dfrac{20.1,31.32\%}{56}=\dfrac{131}{875}\left(mol\right)\\ n_{Ba\left(OH\right)_2}=0,08.1=0,08\left(mol\right)\\ \left[OH^-\right]=\dfrac{\dfrac{131}{875}+0,08.2}{0,02+0,08}=\dfrac{542}{175}\left(M\right)\\ \left[Ba^{2+}\right]=\dfrac{0,08}{0,02+0,08}=0,8\left(M\right)\)

\(\left[K^+\right]=\dfrac{\dfrac{131}{875}}{0,02+0,08}=\dfrac{262}{175}\left(M\right)\)

nH⁺= 0.4*(0.2+0.5)= 0.28 mol

nOH^-= 0.1V + 2*0.2V= 0.5V (mol)

OH^- + H⁺ --> H2O

0.28__0.28

nOH^-= 0.5V= 0.28

=> V= 0.56l=560ml