Đáp án C

n N a 2 S O 4 = 0,02 mol; n_NaCl=0,5 mol; n_Na+= 0,02.2+0,5= 0,54 mol

[Na⁺]= 0,54/(0,117+0,171+0,212)= 1,08M

`n_{BaCl_2}=200.10^{-3}.1=0,2(mol)`

`n_{KCl}=100.10^{-3}.2=0,2(mol)`

`->n_{Cl^-}=2n_{BaCl_2}+n_{KCl}=0,6(mol)`

`->[Cl^-]={0,6}/{(200+100).10^{-3}}=2M`

Bài 1:

Ta có: \(n_{OH^-}=n_{Na^+}=n_{NaOH}=0,2.0,4=0,08\left(mol\right)\)

\(n_{H^+}=n_{Cl^-}=n_{HCl}=0,4.0,3=0,12\left(mol\right)\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

_____0,08_____0,12 (mol)

⇒ n_{OH^- (dư)} = 0,04 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\left[Na^+\right]=\frac{0,08}{0,6}\approx0,133M\\\left[Cl^-\right]=\frac{0,12}{0,6}=0,2M\\\left[OH^-\right]=\frac{0,04}{0,6}\approx0,066M\end{matrix}\right.\)

Câu 2:

Ta có: \(\Sigma n_{K^+}=n_{KCl}+2n_{K_2SO_4}=0,2.1,5+0,3.2.2=1,5\left(mol\right)\)

\(n_{Cl^-}=n_{KCl}=0,2.1,5=0,3\left(mol\right)\)

\(n_{SO_4^{2-}}=0,3.2=0,6\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\left[K^+\right]=\frac{1,5}{0,5}=3M\\\left[Cl^-\right]=\frac{0,3}{0,5}=0,6M\\\left[SO_4^{2-}\right]=\frac{0,6}{0,5}=1,2M\end{matrix}\right.\)

Bạn tham khảo nhé!

Đáp án B

[OH^-]= (0,5.2.0,1+0,1.0,5)/0,2= 0,75M

a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)

\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)

\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)

b, \(\left[Fe^{3+}\right]=\dfrac{\dfrac{2.1,6}{400}}{1,5}\approx0,005M\)

\(\left[K^+\right]=\dfrac{\dfrac{2.6,96}{174}}{1,5}\approx0,053M\)

\(\left[SO_4^{2-}\right]=\dfrac{\dfrac{3.1,6}{400}+\dfrac{6,96}{174}}{1,5}\approx0,035M\)

a) Ta có: \(n_{Al\left(NO_3\right)_3}=\dfrac{4,26}{213}=0,02\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{Al^+}=0,02\left(mol\right)\\n_{NO_3^-}=0,06\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left[Al^+\right]=\dfrac{0,02}{0,1}=0,2\left(M\right)\\\left[NO_3^-\right]=\dfrac{0,06}{0,1}=0,6\left(M\right)\end{matrix}\right.\)

b) Ta có: \(\left[Na^+\right]=0,1+0,02\cdot2+0,3=0,304\left(M\right)\)

c) Bạn xem lại đề !!

Câu 3 : 

\(pH=-log\left[H^+\right]=-log\left(0.1\right)=1\)

Câu 4 : 

Chứa các ion : H⁺ , Cl^-

Câu 5 : 

\(n_{NaOH}=n_{HCl}=0.02\cdot0.1=0.002\left(mol\right)\)

\(\Rightarrow x=\dfrac{0.002}{0.01}=0.2\left(M\right)\)

Câu 1 : 

Bảo toàn điện tích : 

\(n_{SO_4^{2-}}=\dfrac{0.2\cdot2+0.1-0.05}{2}=0.225\left(mol\right)\)

\(m_{Muối}=0.2\cdot64+0.1\cdot39+0.05\cdot35.5+0.225\cdot96=40.075\left(g\right)\)

Câu 2 : 

\(\left[Na^+\right]=\dfrac{0.15\cdot0.5\cdot2+0.05\cdot1}{0.15+0.05}=1\left(M\right)\)

Câu 6 : 

\(pH=14+log\left[OH^-\right]=12\)

\(\Rightarrow\left[OH^-\right]=0.01\)

\(n_{NaOH}=n_{OH}=0.01\cdot2=0.02\left(mol\right)\)

\(\Rightarrow m_{NaOH\left(bđ\right)}=0.02\cdot40=0.8\left(g\right)\)

Câu 7 : 

\(n_{KOH}=0.25\cdot0.03=0.0075\left(mol\right)\)

\(n_{HCl}=0.25\cdot0.01=0.0025\left(mol\right)\)

\(\Rightarrow n_{KOH\left(dư\right)}=0.0075-0.0025=0.005\left(mol\right)\)

\(\left[OH^-\right]=\dfrac{0.005}{0.25+0.25}=0.01\)

\(pH=14+log\left[OH^-\right]=14+log\left(0.01\right)=12\)