CHEM 問題

1.
Let the molecular formula of tribasic acid X be H3A.

H3A + 3NaOH → 3Na3A + 3H2O

No. of moles of NaOH = 0.12 x (27.60/1000) = 0.003312 mol
No. of moles of H3A = 0.003312 x (1/3) = 0.001104 mol
Molar mass of H3A = 0.108/0.001104 = 98 g mol⁻¹


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2.
2NaOH + H2SO4 → Na2SO4 + 2H2O

Volume of H2SO4 = 20.0/1000 = 0.02 dm³
No. of moles of H2SO4 = 1.0 x (20.0/1000) = 0.02 mol
No. of moles of NaOH = 0.02 x 2 = 0.04 mol
Volume of NaOH needed = 0.04/2 = 0.02 dm³

No. of moles of Na2SO4 formed = 0.02 mol
Volume of the final solution = 0.02 + 0.02 dm³ = 0.04 dm³
Concentration of Na2SO4 = 0.02/0.04 = 0.50M


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3.
In A: No. of moles of HCl = 0.05 x (50/1000) = 0.0025 mol
In B: No. of moles of CH3COOH = 0.1 x (25/1000) = 0.0025 mol

1: is incorrect.
HCl is a strong acid which is completely ionized in water to give hydrogenions, but CH3COOH is a weak acid which is partly ionized to givehydrogen ions. Therefore, although theconcentration of HCl and that of CH3COOH are equal, HCl can be completelyionized in water to give a greater number of moles of hydrogen ions.

2: is correct
HCl + NaOH → NaCl + H2O
CH3COOH + NaOH → CH3COONa + H2O
Since the number of moles of HCl and that of CH3COOH are equal. Theyboth need the same number of moles of NaOH for complete neutralization. When0.1 M NaOH is used, both solutions A and B need the same volume of 0.1 M NaOHfor complete neutralization.

2012-05-23 03:33:27 補充：
但係HCL同CH3COOH 既no. of H+ ion 唔同,佢地點解仲要same no. of mole of NaOH既?

HCl is completely ionized to give 0.0025 mol of H^+ ions, which is neutralized by 0.0025 mol of NaOH. ......

2012-05-23 03:33:39 補充：
...... When some NaOH is added to solution B to neutralize some H^+ ions, CH3COOH molecules will be further ionized to compensate the consumption of H^+ ions. When 0.0025 mol of NaOH is added, all CH3COOH is ionized to give 0.0025 mol of H^+ ions which are all neutralized by the NaOH added.

2012-05-23 03:43:50 補充：
如果有一個dibasic acid 既no of mole 同 solution a,b 一樣,咁可唔可以用same volume of NaOH for complete neutralization?

Denote the dibasic acid as H2A.
H2A + 2NaOH → Na2A + 2H2O ......

2012-05-23 03:46:31 補充：
如果有一個dibasic acid 既no of mole 同 solution a,b 一樣,咁可唔可以用same volume of NaOH for complete neutralization?

Denote the dibasic acid as H2A.
H2A + 2NaOH → Na2A + 2H2O

Mole ratio H2A : NaOH = 1 : 2
Doubled volume of NaOH is needed.

1.H3X + 3NaOH à Na3X + 3H2O No. of mole of NaOH= 0.12 x 27.6 / 1000= 3.312 x 10^-3 mol Mole ratio of H3X : NaOH = 1:3 No. of mole of H3X= 3.312 x 10^-3 / 3= 1.104 x 10^-3 mol Molar mass of H3X= 0.108 / (1.104 x 10^-3)= 97.8 g mol^31 2.2NaOH + H2SO4 à Na2SO4 + 2H2O No. of mole of H2SO4= 1 x 20 / 1000= 0.02 mol Mole ratio of NaOH : H2SO4 = 2:1 No. of mole of NaOH= 0.02 x 2= 0.04 mol Volume of NaOH= 0.04 / 2= 0.02 dm^3= 20 cm^3 Mole ratio of H2SO4 : Na2SO4 = 1:1 No. of mole of Na2SO4= 0.02 mol Volume of Na2SO4= 20 + 20= 40 cm^3 Concentration of Na2SO4= 0.02 / 0.04= 0.5 M 3Solution A: 50cm^3 of 0.05M HClSolution B: 25cm^3 of 0.1M CH3COOHWhich of the following statements aboutsolutions A and B is correct?1: A and B have the same number of moles ofH+ (aq) ions in the solution2: A and B require the same volume of 0.1MNaOH for neutralization For 1,Although by calculation the no. of moles ofH+(aq) of the two solutions are the same (2.5 x 10^-3 mol), the actual amountof H+(aq) is not the same. Because HCl is a strong acid while CH3COOHis a weak acidàHCl ionizes completely to formH+(aq); while CH3COOH ionizes slightly to form H+(aq) *that means thereversible arrow is used for ionization of CH3COOH àin HCl, only H+(aq) and Cl-(aq)exist (together with H+ and OH- from water)àin CH3COOH, apart from H+(aq) andCH3COO-(aq), CH3COOH molecules also exist (together with H+ and OH- from water) àthe no. of moles of H+(aq) of HCl ismore than that of CH3COOH For 2,They are both monobasic acidàthey require the same no. of molesof OH-(aq) for neutralizationHCl + NaOH à NaCl + H2OCH3COOH + NaOH à CH3COONa + H2O Therefore, 1 is wrong; 2 is correct.