Chem(Concentration)

For option D,
CH3COOH + NaOH --> CH3COONA + H2O

the number of mole of sodium hydroxide required to react completely with 20 cm^3 of 1 M CH3COOH= (20/1000 X 1) =0.02 mol

2NaOH + H2SO4 --> Na2SO4 + 2H2O

the number of mole of sodium hydroxide required to react completely with 10 cm^3 of 1 M H2SO4 = (10/1000 X 1) X 2= 0.02 mol
Neutralization is irreversible reaction, we don't need to consider the strength of acid or alkali
Thus,option D is correct.

For option A, pH value of acid depends on the number of mole of hydrogen ions.
the pH value of 20 cm^3 of 1 M CH3COOH = -log(0.02)
the pH value of 10 cm^3 of 1 M H2SO4 = -log(0.01x2) =-log(0.2)
However, as ethanoic acid is weak acid which PARTIALLY IONIZE to give a part of hydrogen ions, The concentration of hydrogen ions is actually lower and its real pH value is slightly higher than its theoretic pH value.
As sulphuric acid is strong acid which is able to COMPLETELY IONIZE to give all the hydrogen ions.Its real pH value is the same as its theoretic pH value.

For options B & C, similarly, the concentration of hydrogen ions in sulphuric acid is higher, it has better electrical conductivity and reacts with Mg at a higher rate.

Thus, option A,B,C are all incorrect.

因為complete neutralization 是要令acid molecule上面的所有可電離的H都比neutralize,無須理會最後pH是幾多, 只是當他們全部都會被反應
CH3COOH只有一個可以電離的H, 在-COOH上, 最後變成CH3COO-, 而H2SO4有兩個可以電離的H, 最後變成SO42-,
這兩個都是同樣濃度, 但乙酸的volume是硫酸的兩倍,
因此molecule 的數量 是乙酸是硫酸的兩倍,
而硫酸可以被反應的H的數量是乙酸的兩倍
因此他們需要的NaOH是一樣

講深D, 就是他電離是一個平衡,
電離同重合的反應(向前和向後)是同時進行, 但他們的速度取決於兩邊的濃度
弱酸電離後的H+好快就被OH-食左, 返吾到轉頭, 沒有H+可以同個anion重合, 所以總的就是繼續電離
這個其實都是比喻