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

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

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

Ta có: \(n_{H^+}=2n_{H_2SO_4}+n_{HCl}=2\cdot0,02+0,07=0,11\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,11}{0,35+0,05}=0,275\left(M\right)\) \(\Rightarrow\left[OH^-\right]=\dfrac{10^{-14}}{\left[H^+\right]}\approx3,64\cdot10^{-14}\left(M\right)\)

\(\Rightarrow pH=-log\left(0,275\right)\approx0,56\)

*Môi trường axit và làm quỳ tím hóa đỏ 

\(n_{H_2SO_4}=0,05.0,4=0,02\left(mol\right)\\ n_{HCl}=0,2.0,35=0,07\left(mol\right)\\ \left[H^+\right]=\dfrac{0,02.2+0,07}{0,05+0,35}=0,275\left(M\right)\\ pH=-log\left[H^+\right]=-log\left[0,275\right]=0,56\\ \Rightarrow Qùy.hóa.đỏ\)

\(n_{HCl}=0.1\cdot0.03=0.003\left(mol\right)\)

\(n_{NaOH}=0.1\cdot0.01=0.001\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

Lập tỉ lệ : 

\(\dfrac{0.003}{1}>\dfrac{0.001}{1}\Rightarrow HCldư\)

\(n_{HCl\left(dư\right)}=0.003-0.001=0.002\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.002}{0.1+0.1}=0.01\)

\(pH=-log\left(0.01\right)=2\)

\(b.\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(0.001..........0.002\)

\(V_{Ba\left(OH\right)_2}=\dfrac{0.001}{1}=0.001\left(l\right)\)

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

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

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

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 :(

a) ko bik ....

b) 13,5

c)2

d)2,4

C_{M ( A)}=0.325

C_{M (B)=0.225}