\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)

\(n_{HCl}=0.1\cdot0.5=0.05\left(mol\right)\)

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

\(0.05.......0.05.......0.05\)

Dung dịch D : 0.05 (mol) KOH , 0.05 (mol) KCl

\(\left[K^+\right]=\dfrac{0.05+0.05}{0.1+0.1}=0.5\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(\left[OH^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

\(0.05.........0.025\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.025}{1}=0.025\left(l\right)\)

a) \(n_{NaOH}=0,2.1=0,2\left(mol\right)\); \(n_{HNO_3}=0,2.0,5=0,1\left(mol\right)\)

\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)

0,2.............0,1

Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\) => Sau phản ứng NaOH dư

Dung dịch D gồm NaNO3 và NaOH dư

\(n_{NaNO_3}=n_{HNO_3}=0,1\left(mol\right)\)

\(n_{NaOH\left(pứ\right)}=n_{HNO_3}=0,1\left(mol\right)\)

\(n_{NaOH\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

Ion trong dung dịch D : Na+ , NO3-, OH-

\(\left[Na^+\right]=\dfrac{0,1+0,1}{0,2}=1M\)

\(\left[NO_3^-\right]=\dfrac{0,1}{0,2}=0,5M\)

\(\left[OH^-\right]=\dfrac{0,1}{0,2}=0,5M\)

b)Trong dung dịch D chỉ có NaOH dư phản ứng

 \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

0,1................0,05

=> \(V_{H_2SO_4}=\dfrac{0,05}{1}=0,05\left(l\right)\)

Tách ra lần bài đi em !

[K+]=[Cl-]=0,25M

[KOH dư]=0,25M

b)   2KCl   +    H₂SO₄    ----------->K₂SO₄   + 2HCl

  0,05(mol)---->0,025(mol)

=>v_H2SO4=\(\frac{0,025}{1}\)=0,025(lít)

c)pH=-log(0,25)=0,602

(câu c mình không chắc chắn lắm nha bạn!!!)

câu a) làm sao vậy mn chỉ mik vs

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

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

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

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

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

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

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

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

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

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

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

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

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

\(b.\)

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

\(c.\)

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

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

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

\(a.\)

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

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

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

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

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

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

\(b.\)

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

\(c.\)

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

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

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

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

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

\(n_{NaOH}=0.25\cdot2=0.5\left(mol\right)\)

\(n_{H_2SO_4}=0.25\cdot1=0.25\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)

\(0.5..............0.25................0.25\)

\(\left[Na^+\right]=\dfrac{0.25\cdot2}{0.25+0.25}=1\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.25}{0.25+0.25}=0.5\left(M\right)\)

a) \([OH^-]=\left[KOH\right]=1,5M\)

b) Để trung hòa dung dịch A: \(n_{\left[OH^-\right]}=n_{\left[H^+\right]}\)

                                          \(\Rightarrow0,15=0,5\cdot V_{HCl}\Rightarrow V_{HCl}=0,3l=300ml\)