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
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
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!
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
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.