\(n_{NaOH}=0.015\cdot2=0.03\left(mol\right)\)

\(n_{H_2SO_4}=0.015\cdot1.5=0.0225\left(mol\right)\)

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

\(0.03..........0.015...........0.015\)

\(n_{H_2SO_4\left(dư\right)}=0.0225-0.015=0.0075\left(mol\right)\)

\(C_{M_{Na^+}}=\dfrac{0.015\cdot2}{0.015+0.015}=1\left(M\right)\)

\(C_{M_{H^+}}=\dfrac{0.0075\cdot2}{0.015+0.015}=0.5\left(M\right)\)

\(C_{M_{SO_4^{2-}}}=\dfrac{0.015+0.0075}{0.015+0.015}=0.75\left(M\right)\)

\(\left[Na^+\right]=\dfrac{2.0,015}{2.0,015}=1M\)

\(\left[OH^-\right]=\dfrac{2.0,015}{2.0,015}=1M\)

\(\left[H^+\right]=\dfrac{2.1,5.0,015}{2.0,015}=1,5M\)

\(\left[SO_4^{2-}\right]=\dfrac{1,5.0,015}{2.0,015}=0,75M\)

nNaOH=0,025mol

nH2SO4=0,015mol

2NaOH+H2SO4->Na2SO4+2H2O

Ta có 0,025/2 <0,015/1 =>H2SO4 dư

Khi nhúng quì tím vào dd thì quì tím chuyển sang màu đỏ

2NaOH+H2SO4->Na2SO4+2H2O

0,025     0,0125      0,0125

DD X: H2SO4:0,0025mol

Na2SO4: 0,0125mol

C(H2SO4)=0,00625M

C(NaOH)=0,03125M

3. Cần dùng KOH 1M để trung hòa dd X

H2SO4+2KOH->K2SO4+H2O

0,0025    0,005

V(KOH)=n/C=0,005lit=5ml

1) $n_{NaOH} = 0,015(mol) ; n_{H_2SO_4} = 0,025(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + H_2O$

Ta thấy : 

$n_{NaOH} : 2 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư

Do đó quỳ tím hóa đỏ.

2)

$n_{Na_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,0075(mol)$
$n_{H_2SO_4\ dư} = 0,025 - 0,0075 = 0,0175(mol)$

$V_{dd\ X} = 0,15 + 0,25 = 0,4(lít)$

Suy ra : 

$C_{M_{Na_2SO_4}} = \dfrac{0,0075}{0,4} = 0,01875M$
$C_{M_{H_2SO_4\ dư}} = \dfrac{0,0175}{0,4}  = 0,04375M$

3)

$2KOH + H_2SO_4 \to K_2SO_4 + H_2O$
$n_{KOH} = 2n_{H_2SO_4\ dư} = 0,035(mol)$
$V_{dd\ KOH} =\dfrac{0,035}{1} = 0,035(lít)$

\(n_{NaOH}=0,015.2=0,03\left(mol\right)\)

\(n_{H_2SO_4}=0,01.1,5=0,015\left(mol\right)\)

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

Xét: \(\frac{n_{NaOH}}{2}=0,015=\frac{n_{H_2SO_4}}{1}=0,015\)\

=> NaOH và H₂SO₄ vừa đủ

=> Dung dịch sau phản ứng trung hòa

a, \(n_{KOH}=0,15.2=0,3\left(mol\right);n_{H_2SO_4}=0,25.2=0,5\left(mol\right)\)

PTHH: 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O

Mol:       0,3          0,15         0,15

Ta có: \(\dfrac{0,3}{2}< \dfrac{0,5}{1}\) ⇒ KOH hết, H₂SO₄ dư

  ⇒ dd sau pứ làm quỳ tím chuyển thành màu đỏ

b, \(C_{M_{ddH_2SO_4}}=\dfrac{0,5-0,15}{0,15+0,25}=0,875M\) 

 \(C_{M_{ddK_2SO_4}}=\dfrac{0,15}{0,15+0,25}=0,375M\)

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

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

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

\(n_{NaOH}=\dfrac{12}{40}=0.3\left(mol\right)\)

\(n_{H_2SO_4}=0.1\cdot1=0.1\left(mol\right)\)

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

\(0.2..............0.1................0.1\)

\(m_A=m_{Na_2SO_4}+m_{NaOH\left(dư\right)}=0.1\cdot142+\left(0.3-0.2\right)\cdot40=18.2\left(g\right)\)