Chem - PH Question
Calculate the concentration, in mol dm–3, of an aqueous solution of sulfuric acid that has a pH of 0.25
the mark scheme says:
[H+] = 0.56
[H2SO4] = ½ × 0.56 = 0.28
I understand how they got [H+]. However I dont understand why there dividing the answer by two? Are you not meant to times by 2. please kindly explain
the mark scheme says:
[H+] = 0.56
[H2SO4] = ½ × 0.56 = 0.28
I understand how they got [H+]. However I dont understand why there dividing the answer by two? Are you not meant to times by 2. please kindly explain
Original post by ZZZ_A
Calculate the concentration, in mol dm–3, of an aqueous solution of sulfuric acid that has a pH of 0.25
the mark scheme says:
[H+] = 0.56
[H2SO4] = ½ × 0.56 = 0.28
I understand how they got [H+]. However I dont understand why there dividing the answer by two? Are you not meant to times by 2. please kindly explain
the mark scheme says:
[H+] = 0.56
[H2SO4] = ½ × 0.56 = 0.28
I understand how they got [H+]. However I dont understand why there dividing the answer by two? Are you not meant to times by 2. please kindly explain
essentially the concentration of protons is a measure of the concentration of the H2SO4. As there are 2 protons dissociated for each H2SO4 you need to divide the concentration of H+ by 2. This is only true for strong acids that can completely ionise though!
Original post by ZZZ_A
Calculate the concentration, in mol dm–3, of an aqueous solution of sulfuric acid that has a pH of 0.25
the mark scheme says:
[H+] = 0.56
[H2SO4] = ½ × 0.56 = 0.28
I understand how they got [H+]. However I dont understand why there dividing the answer by two? Are you not meant to times by 2. please kindly explain
the mark scheme says:
[H+] = 0.56
[H2SO4] = ½ × 0.56 = 0.28
I understand how they got [H+]. However I dont understand why there dividing the answer by two? Are you not meant to times by 2. please kindly explain
Do you certainly know that protons will be dissociated in a solution and H+ ions remain. The more H-atoms an acid has in the bond, the more can be dissociated and thus ionised. The PH value 0.56 is the one for a completely dissociation of sulphuric acid in a solution, so for two H-atoms ionised to a proton. As it is needed the concentration of one dissociated H-atom in the calculation, you need to divide the value by two.
@TypicalNerd correct me, if I am wrong.
What they want you to recognise is that pH = -log[H^+] and so rearranging and solving gives you [H^+].
However, at A level, they (incorrectly) expect you to assume that both the hydrogens on each H2SO4 dissociate off the molecule and so you have
H2SO4 —> SO4^2- + 2H^+
Notice from this how for every one H2SO4 you have two H^+ ions? So [H^+] = 2[H2SO4] and thus 1/2 [H^+] = [H2SO4]
However, at A level, they (incorrectly) expect you to assume that both the hydrogens on each H2SO4 dissociate off the molecule and so you have
H2SO4 —> SO4^2- + 2H^+
Notice from this how for every one H2SO4 you have two H^+ ions? So [H^+] = 2[H2SO4] and thus 1/2 [H^+] = [H2SO4]
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