a) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

b)

n H2SO4 = 0,03.1 = 0,03(mol)

n NaOH = 2n H2SO4 = 0,06(mol)

=> CM NaOH = 0,06/0,05 = 1,2M

c) $H_2SO_4 + 2KOH \to K_2SO_4 + 2H_2O$

n KOH = 2n H2SO4 = 0,06(mol)

=> m KOH = 0,06.56 = 3,36 gam

=> m dd KOH = 3,36/5,6% = 60(gam)

=> V dd KOH = m/D = 60/1,045 = 57,42(ml)

NaOH+CH₃COOH->CH₃COONa+H₂O

0,3----------0,3 mol

m CH3COOH= 18g

=>n _CH3COOH=\(\dfrac{18}{60}\)=0,3 mol

V_NaOH=\(\dfrac{0,3}{1}\)=0,3l=300ml

=>C

Ta có: \(n_{Ba\left(OH\right)_2}=0,025.0,02=0,0005\left(mol\right)\)

\(n_{NaOH}=0,025.0,05=0,00125\left(mol\right)\)

\(\Rightarrow n_{OH^-}=2n_{Ba\left(OH\right)_2}+n_{NaOH}=0,00225\left(mol\right)\)

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

\(\Rightarrow n_{H^+}=n_{OH^-}=0,00225\left(mol\right)\)

Gọi: V_X = x (l)

Ta có: \(n_{HCl}=0,1x\left(mol\right)\)

\(n_{CH_3COOH}=0,2x\left(mol\right)\)

\(\Rightarrow n_{H^+}=n_{HCl}+n_{CH_3COOH}=0,1x+0,2x=0,00225\)

\(\Rightarrow x=0,0075\left(l\right)=7,5\left(ml\right)\)

Tham khảo 

để thu được muối photphat trung hòa , cần lấy bao nhiêu ml dung dịch NaOH 1M cho tác dụng với 50ml dung dịch H3PO4 0,5M ? - Hoc24

\(n_{H_3PO_4}=0.05\cdot0.5=0.025\left(mol\right)\)

\(3NaOH+H_3PO_4\rightarrow Na_3PO_4+3H_2O\)

\(0.075......0.025\)

\(V_{dd_{NaOH}}=\dfrac{0.075}{1}=0.075\left(l\right)=75\left(ml\right)\)

ĐÁP ÁN C

Trung hòa khi n_CH3COOH=n_NaOH => V= 200ml => chọn C

$n_{H_3PO_4} = 0,05.0,5 = 0,025(mol)$
$3NaOH + H_3PO_4 \to Na_3PO_4 + 3H_2O$

Theo PTHH : 

$n_{NaOH} = 3n_{H_3PO_4} = 0,075(mol)$
$V_{dd\ NaOH} = \dfrac{0,075}{1} = 0,075(lít) = 75(ml)$

\(n_{NaOH}=0.05\cdot1=0.05\left(mol\right)\)

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

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

\(m_{dd_{H_2SO_4}}=\dfrac{0.025\cdot98}{19.6\%}=12.5\left(g\right)\)