F.4 Chem MC *40marks* with explanation THX THX THX

(1) Please note that in the question, it mentions "acidic" solution instead of "acid". Therefore, a copper(II) sulphate acidified with sulphuric acid can also be considered as an "acidic" solution.
However, when acidified copper(II) sulphate solution undergoes electrolysis, owing to the lower position of copper than hydrogen in ECS, copper(II) ions are preferentially discharged and hence NO gas bubbles are given out at the cathode.
So option IV is not necessarily inc orrect.
For options II and III, we may use the fact that H+ ions are present with more amount in acidic solution to explain and option I can be explained based on the definition.
So ans = B.
(2) Looking at the equation:
H2SO4 + 2KOH → K2SO4 + 2H2O
We can see that 2 moles of KOH completely neutralize 1 mole of H2SO4. Hence, volume of KOH needed = 200 cm3.
Then, resulting solution = 300 cm3
No. of moles of salt (K2SO4) formed = 0.12 moles (no. of moles sulphuric acid originally)
Hence, molarity of salt in resulting solution = 0.4 M
Ans = A
(3) Ans = A
Suppose that the liquid is chlorine water (bleach), if all aqueous part (water) is blown out, those remained are salt of sodium hypochlorite (active ingredient of bleach) and no chlorine gas will be evolved.
(4) Ans = D
In fact, for option A, the solution looks very pale yellow and still it can be decolourized if blown into sodium sulphite solution. However, human visual systems are not that sensitivie to detect the colour change. Therefore it is regarded as wrong answer.
For option B, since chlorine and HCl are both acidic in aqueous state, so pH test fails to distinguish the presence.
For option C, potassium dichromate solution is also an oxidizing agent that cannot react with chlorine gas which is an oxidizing agent too.
Finally for option D, iodide ions are readily oxidized by chlorine to give iodine:
2I- + Cl2 → I2 + 2Cl-
And further, those iodine formed dissolves in potassium iodide solution, giving a brown complex:
I- + I2 → I3-

1.解決方案可能被考慮如產生酸的如果它
我。 轉藍色的石蕊紅色。
2. 當和鈉碳酸鹽混合，解放一種無顏色的瓦斯。
3. 以鎂冒泡。
4. 接受電分解，在陰極解放另一種無顏色的瓦斯。
（一） 我和 3 只有。
（b） 我， 2 和 3 只有。
（c） 2， 3 和 4 只有。
（d） 我， 2， 3 和 4