Đáp án D

a, \(n_{HCl}=0,1.0,2=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)

\(n_{H_2SO_4}=0,1.0,2=0,02\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,04\left(mol\right)\)

\(n_{NaOH}=0,3.0,4=0,12\left(mol\right)=n_{Na^+}=n_{OH^-}\)

\(\Rightarrow\sum n_{H^+}=0,02+0,04=0,06\left(mol\right)\)

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

0,06__0,06 (mol)

⇒ n_OH^- dư = 0,12 - 0,06 = 0,06 (mol)

\(\Rightarrow\left\{{}\begin{matrix}\left[Cl^-\right]=\dfrac{0,02}{0,1+0,3}=0,05\left(M\right)\\\left[SO_4^{2-}\right]=\dfrac{0,02}{0,1+0,3}=0,05\left(M\right)\\\left[Na^+\right]=\dfrac{0,12}{0,1+0,3}=0,3\left(M\right)\\\left[OH^-\right]=\dfrac{0,06}{0,1+0,3}=0,15\left(M\right)\end{matrix}\right.\)

b, pH = 14 - (-log[OH^-]) ≃ 13,176

Đáp án : A

dd X : n_H+ = 2n_H2SO4 + n_HCl = 0,02 mol

dd Y : n_OH = n_NaOH + 2n_Ba(OH)2 = 0,04 mol

=> Trong Y : n_OH – n_H+ = 0,02 mol = n_OH- => C_OH = 0,1M => pH = 13

a) \(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,1\cdot0,12+0,1\cdot0,04=0,016\)

    \(C_M=\dfrac{0,016}{0,2}=0,08M\)

     \(\Rightarrow pH=-log\left(0,08\right)=1,1\)

b) \(n_{OH^-}=n_{KOH}+2n_{Ba\left(OH\right)_2}=0,012+2\cdot0,004=0,02\)

     \(C_M=\dfrac{0,02}{0,2}=0,1\)

     \(\Rightarrow pH=-log\left(\dfrac{10^{-14}}{0,1}\right)=13\)

Câu 2 : 

$n_{HCl} = 0,2.1 + 0,3.1,5 = 0,65(mol)$
$V_{dd} = 0,2 + 0,3 = 0,5(mol)$
$C_{M_{HCl}} = \dfrac{0,65}{0,5} =1,3M$

Câu 3 : 

Gọi $m_{H_2O\ cần\ thêm} =a (gam)$

Sau khi thêm : 

$m_{NaOH} = 100.35\% = 35(gam)$
$m_{dd} = 100 + a(gam)$

Suy ra:  $\dfrac{35}{100 + a}.100\% = 20\%$

Suy ra: a = 75(gam)

Câu 4 : 

Gọi $V_{dd\ HCl\ 2M} =a (lít) ; V_{dd\ HCl\ 3M} = b(lít)$

Ta có : 

$a + b = 4$
$2a + 3b = 4.2,75$

Suy ra a = 1(lít) ; b = 3(lít)