Đáp án B

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

\(n_{NaOH}=0.1\cdot0.1=0.01\left(mol\right)\)

\(n_{KOH}=0.1\cdot0.1=0.01\left(mol\right)\)

\(V=0.1+0.1=0.2\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)

\(\left[K^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)

\(\left[OH^-\right]=\dfrac{0.01+0.01}{0.2}=0.1\left(M\right)\)

\(b.\)

\(pH=14+log\left[OH^-\right]=14+log\left(0.1\right)=13\)

\(c.\)

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

\(0.02........0.02\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.02}{1}=0.02\left(l\right)\)

\(a.\)

\(n_{NaOH}=0.1\cdot0.1=0.01\left(mol\right)\)

\(n_{KOH}=0.1\cdot0.1=0.01\left(mol\right)\)

\(V=0.1+0.1=0.2\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)

\(\left[K^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)

\(\left[OH^+\right]=\dfrac{0.01+0.01}{0.2}=0.1\left(M\right)\)

\(b.\)

\(pH=14+log\left(0.1\right)=13\)

\(c.\)

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

\(0.02.......0.02\)

\(V_{H_2SO_4}=\dfrac{0.02}{1}=0.02\left(l\right)\)

a) Ta có: \(n_{NaOH}=0,1\cdot0,1=n_{KOH}=0,01\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{OH^-}=0,02\left(mol\right)\\n_{Na^+}=n_{K^+}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[OH^-\right]=\dfrac{0,02}{0,2}=0,1\left(M\right)\\\left[Na^+\right]=\left[K^+\right]=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)

b) Ta có: \(pH=14+log\left[OH^-\right]=13\)

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

Theo PT ion: \(n_{H^+}=n_{OH^-}=0,02\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4}=0,01\left(mol\right)\) \(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,01}{1}=0,01\left(l\right)=10\left(ml\right)\)

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

     \(C_M=\dfrac{0,02}{0,2}=0,1\)

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

a)

$KOH + HCl \to KCl + H_2O$

$n_{KOH} = 0,3(mol) < n_{HCl} = 1,05(mol)$ nên HCl dư

$n_{HCl\ dư} = 1,05 -0 ,3 = 0,75(mol)$
$n_{KCl} = n_{KOH} = 0,3(mol)$

$V_{dd} = 0,3+ 0,7 = 1(lít)$
Suy ra : 

$[K^+] = \dfrac{0,3}{1} = 0,3M$
$[Cl^-] = \dfrac{0,75 + 0,3}{1} = 1,05M$
$[H^+] = \dfrac{0,75}{1} = 0,75M$

b)

$Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{Ba(OH)_2} = \dfrac{1}{2}n_{HCl} = 0,375(mol)$

$V_{Ba(OH)_2} = \dfrac{0,375}{1,5} = 0,25(lít)$

\(n_{KOH}=0.3\cdot1=0.3\left(mol\right)\)

\(n_{HCl}=0.7\cdot1.5=1.05\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

\(0.3...........0.3..........0.3\)

Dung dịch D gồm : 0.3 (mol) KCl , 0.75 (mol) HCl dư

\(\left[K^+\right]=\dfrac{0.3}{0.3+0.7}=0.3\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.3+0.75}{0.3+0.7}=1.05\left(M\right)\)

\(\left[H^+\right]=\dfrac{0.75}{0.3+0.7}=0.75\left(M\right)\)

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

\(0.375..................0.75\)

\(V_{dd_{Ba\left(OH\right)_2}}=\dfrac{0.375}{1.5}=0.25\left(l\right)\)

Gọi \(\left\{{}\begin{matrix}C_{M\left(A\right)}=aM\\C_{M\left(B\right)}=bM\end{matrix}\right.\)

Giả sử trộn 50ml dd A với 50ml dd B để thu được 100ml dd C

=> \(\left\{{}\begin{matrix}n_{NaOH}=0,05a\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,05b\left(mol\right)\end{matrix}\right.\)

\(n_{BaSO_4}=\dfrac{9,32}{233}=0,04\left(mol\right)\)

n_H2SO4 = 0,035.2 = 0,07 (mol)

PTHH: Ba(OH)₂ + H₂SO₄ --> BaSO₄ + 2H₂O

              0,04<----0,04<-------0,04

            2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

              0,06<----0,03

=> \(\left\{{}\begin{matrix}0,05a=0,06\\0,05b=0,04\end{matrix}\right.\)

=> a = 1,2; b = 0,8 

20 ml dd A chứa n_NaOH = 0,02.1,2 = 0,024 (mol)

\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

PTHH: 2NaOH + Al₂O₃ --> 2NaAlO₂ + H₂O

             0,024-->0,012

            Ba(OH)₂ + Al₂O₃ --> Ba(AlO₂)₂ + H₂O

             0,188<---0,188

=> \(V_{dd.B}=\dfrac{0,188}{0,8}=0,235\left(l\right)=235\left(ml\right)\)