✔ 最佳答案
The answer is C.
No. of moles of NaOH used = MV = 0.1 x (10/1000) = 0.001 mol
A.
CH3COOH + NaOH → CH3COONa + H2O
Mole ratio CH3COOH : NaOH = 1 : 1
No. of moles of CH3COOH used = 0.001 mol
Volume of CH3COOH = mol/M = 0.001/0.15 = 0.00667 dm3
B.
H2SO4 + 2NaOH → Na2SO4 + 2H2O
Mole ratio H2SO4 : NaOH = 1 : 2
No. of moles of H2SO4 = 0.001 x (1/2) = 0.0005 mol
Volume of H2SO4 = mol/M = 0.0005/0.1 = 0.005 dm3
C.
H3PO4 + 3NaOH → Na3PO4 + 3H2O
Mole ratio H3PO4 : NaOH = 1 : 3
No. of moles of H3PO4 = 0.001 x (1/2) = 0.000333 mol
Volume of H3PO4 = mol/M = 0.000333/0.1 = 0.00333 dm3
(smallest volume)
D.
HCl + NaOH → NaCl + H2O
Mole ratio HCl : NaOH = 1 : 1
No. of moles of HCl = 0.001 mol
Volume of HCl = mol/M = 0.001/0.2 = 0.005 dm3
