Đáp án:

`a)` `CM_A= 0,7M, \ CM_B=1,1M`

`b)``V_B: V_A=` \(\left[ \begin{array}{l}10:1\\16,5:1\end{array} \right.\) 

Giải thích các bước giải:

`a)` Cho $\begin{cases} H_2SO_4: xM\\ NaOH: yM\\\end{cases}$

`2NaOH+H_2SO_4\to Na_2SO_4+2H_2O`

`TN_1:` $\begin{cases} n_{H_2SO_4}=0,2x(mol)\\ n_{NaOH}=0,3y(mol)\\\end{cases}$

Do `C` làm quỳ tím hóa xanh `=>` `NaOH` phản ứng dư

`NaOH+HCl\to NaCl+H_2O`

`n_{HCl}=0,05.0,04=0,002(mol)`

`n_{NaOH \text{20ml}}=n_{HCl}=0,002(mol)`

`n_{NaOH\text{dư}}=\frac{0,002.1000.0,5}{20}=0,05(mol)`

`n_{NaOH \text{dư}}=0,3y-0,4x=0,05(mol)`

`=> n_{NaOH \text{dư}}=-0,4x+0,3y=0,05(mol)(1)`

`TN_2:` $\begin{cases} n_{H_2SO_4}=0,3x(mol)\\n_{NaOH}=0,2y(mol)\\\end{cases}$

Do `D` làm quỳ tím hóa đỏ `=>` `H_2SO_4` dư.

`n_{H_2SO_4\text{dư}}=0,3x-0,1y(mol)`

`H_2SO_4+2NaOH\to Na_2SO_4+2H_2O`

`n_{NaOH }=0,08.0,1=0,008(mol)`

`=> n_{H_2SO_4\text{ 20ml}}=\frac{0,008}{2}=0,004(mol)`

`=> n_{H_2SO_4\text{ dư}}=\frac{0,004.1000.0,5}{20}=0,1(mol)`

`=>n_{H_2SO_4\text{dư}}=3x-0,1y=0,1(mol)(2)`

`(1),(2)=> x=CM_{H_2SO_4}=CM_A=0,7M`

`y=CM_{NaOH}=CM_B=1,1M`

`b)` `E` chứa `H_2SO_4` và `NaOH`

`n_{BaCl_2}=0,1.0,15=0,015(mol)`

`n_{BaSO_4}=\frac{3,262}{233}=0,014(mol)`

`n_{Al_2O_3}=\frac{3,262}{102}\approx0,032(mol)`

`BaCl_2+H_2SO_4\to BaSO_4\downarrow +2HCl`

Do `n_{BaCl_2}>n_{BaSO_4}`

`=>` `BaCl_2` dư.

`=> n_{H_2SO_4}=0,014(mol)`

`V_A=\frac{0,014}{0,7}=0,02(l)`

Do `AlCl_3` tác dụng với `NaOH` nên ta xét `2` trường hợp:

`TH_1:` `NaOH` phản ứng hết với muối `AlCl_3`

`n_{AlCl_3}=0,1.1=0,1(mol)`

`3NaOH+AlCl_3\to 3NaCl+Al(OH)_3`

`2Al(OH)_3\overset{t^o}{\to}Al_2O_3+3H_2O`

Do `n_{Al(OH)_3}<n_{AlCl_3}`

`=> ` `AlCl_3` dư

`n_{NaOH \text{ pứ trung hòa với acid}}=0,028(mol)`

`=>∑n_{NaOH}=0,028+3.0,064=0,22(mol)`

`=> V_{B}=\frac{0,22}{1,1}=0,2(mol)`

`=> \frac{V_B}{V_A}=\frac{0,2}{0,02}=\frac{10}{1}=10:1`

`TH_2`: `NaOH` phản ứng dư.

`NaOH+Al(OH)_3\to NaAlO_2+2H_2O`

`∑n_{NaOH}=0,3.1+0,032.2=0,364(mol)`

`=> V_B=\frac{0,364}{1,1}\approx 0,33(l)`

`=> \frac{V_B}{V_A}=\frac{0,33}{0,02}=16,5:1`

Vậy `V_B: V_A=` \(\left[ \begin{array}{l}10:1\\16,5:1\end{array} \right.\)