Đáp án C

n_H+ = 0,02 mol; n_OH-= 0,04 mol

H⁺    + OH^- → H₂O

0,02   0,04

n_{OH- dư} = 0,02 mol; [OH^-] _dư = 0,02/0,2 = 0,1M suy ra [H⁺] = 10^-13 M suy ra pH = 13

Yêu cầu đề bài?

Đáp án B

 [H⁺]= 0,01.2+0,05= 0,07M nên pH= -log[H⁺]=1,15

\(n_{H^+}=0.2\cdot0.01\cdot2=0.004\left(mol\right)\)

\(n_{OH^-}=0.1\cdot0.01=0.001\left(mol\right)\)

\(H^++OH^-\rightarrow H_2O\)

\(0.001.....0.001\)

\(n_{H^+\left(dư\right)}=0.004-0.001=0.003\left(mol\right)\)

\(pH=-log\left(H^+\right)=-log\left(\dfrac{0.003}{0.2+0.1}\right)=2\)

Có: \(n_{H^+}=2n_{H_2SO_4}=2.0,2.0,01=0,004\left(mol\right)\)

\(n_{OH^-}=0,1.0,01=0,001\left(mol\right)\)

PT ion: \(H^++OH^-\rightarrow H_2O\)

____0,004___0,001 (mol)

\(\Rightarrow n_{H^+\left(dư\right)}=0,003\left(mol\right)\)\\(\Rightarrow\left[H^+\right]=\dfrac{0,003}{0,3}=0,01\)

\(\Rightarrow pH=2\)

Bạn tham khảo nhé!

\(H^++OH^-\rightarrow H_2O\\ n_{H^+}=0,05\left(mol\right);n_{OH^-}=0,07\left(mol\right)\\ Lậptỉlệ:\dfrac{0,05}{1}< \dfrac{0,07}{1}\\ \Rightarrow OH^-dư\\ \left[OH^-_{dư}\right]=\dfrac{0,07-0,05}{0,2}=0,1M\\ \Rightarrow pOH=-log\left(0,1\right)=1\\ \Rightarrow pH=14-1=13\)