Acids and Bases

2008-06-25 3:11 am

Which of the folowing acid solutions would require the smallest volume to completely neutralise 10mL of 0.100M NaOH?
A. 0.150M CH3COOH
B. 0.100M H2SO4
C. 0.100M H3PO4
D. 0.200M HCl

求詳盡答案

更新1:

Should there be different in calculation between weak acid and strong acid?

回答 (1)

Uncle Michael

2008-06-25 7:02 am

✔ 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

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