f6 化學問題2題 help!!!!!!!!

Question 1:
The data given in the question are probably wrong. Please check again.

Basicity of an acid refers to the number of replaceable hydrogen atoms in one molecule of the acid. It can have a value of 1, 2, 3 or 4.

For basicity = 1, NaOH + HX -> salt + water
For basicity = 2, 2NaOH + H2X -> salt + water
For basicity = 3, 3NaOH + H3X -> salt + water
For basicity = 4, 4NaOH + H4X -> salt + water

Basicity of the acid = ratio of number of moles of base to number of moles of acid
Basicity of the acid = number of moles of NaOH/number of moles of acid
(0.1*23.4/1000)/[(2.3/197)*(10/250)]
0.00234/0.000467 = 5.01 (wrong)

Molarity mass of Acetylsalicylic acid is 197. (Aspirin) HC9H7O4. The acid is a monoprotic acid ~ an acid that can donate one proton. (basicity = 1). Your given answer B 2 is wrong.


Question 2:

2 KOH + H2SO4 -> K2SO4 + H2O
2 moles of KOH reacts with 1 moles of H2SO4
Number of moles in 20cm3 of 2M KOH = 2/1000 x 20 = 0.04 mol
Number of moles in 25cm3 of 1M H2SO4 = 1/1000 x 25 = 0.025 mol
0.04 moles of KOH reacts with 0.02 moles of H2SO4
The unreacted H2SO4 = 0.025 mol – 0.02 mol = 0.005 moles
The total volume of solution = 20cm3 + 25 cm3 = 45 cm3
That is 0.005 moles in 45 mL solution
The molarity of unreacted H2SO4 = 0.005 moles/45 x 1000 = 0.1111 M
H2SO4 -> 2[H]+ and SO4-20.1111M H2SO4 dissociates into 0.2222 M [H+]
pH = -log[H+] = -log[0.2222] = 0.6532

The answer is C 0.65

1. Is the molar mass of the acid given? If not, only the following relationship can be listed......

Let n be the basicity of the acid
(0.1*23.4/1000)/[(2.3/molar mass)*(10/250)] = 1/n,
mol of NaOH : mol of acid = 1:n

2. The answer I have found is B (pH = 2), though I'm not sure whether I've made any mistake......

2 KOH + H2SO4 -> H2O + K2SO4
Mol of KOH = 2*20/1000 = 0.04 mol
Mol of H2SO4 = 1*25/1000 = 0.025 mol
Since KOH : H2SO4 = 2:1, therefore mol of KOH reacted = 0.02 mol
Resultant mol of H2SO4 = 0.025-0.02 = 0.005 mol
1 mol of H2SO4 can ionize to give 2H+, thus pH = -log(0.005*2) = 2