A 0.552g sample of ascorbic acid was dissolved in water to a totalvolume of 20mL

Question

A 0.552g sample of ascorbic acid was dissolved in water to a totalvolume of 20mL and titrated with a .1103 M KOH and the equivalencepoint occured when 28.42 mL of base was added. The pH of thesolution at 10mL of base added was 3.72

Thanks Much!

Explanation / Answer

Ascorbic acid is HC6H7O6 First write a balanced chemical equation:     HC6H7O6 +OH- -->C6H7O6- +H2O Then calculate the moles of base added at the equivalence point(this is the point where moles of acid = moles of base):     n(KOH) = 0.1103 mol/L * 0.02842 L =3.135x10-3 mol Since the reaction is 1-to-1, this means we had3.135x10-3 mol of ascorbic acid at the start. We alsoknow that 0.552 g was used, so we can easily calculate the molarmass:     MW(acid) = 0.552 g / 3.135x10-3 mol = 176.1g/mol (You can check this on the net - it's correct) I don't think my answer for the next part is correct, but here'sthe working: At equilibrium, there will be no more OH- remaining, asthis will have been used to neutralise the ascorbic acid. Thismeans the equilibrium involves only ascorbic acid::     HC6H7O6 H+ +C6H7O6- For calculating the acid dissociation constant, remember that:     Ka =[H+][A-]/[HA] Since the equilibrium is 1-to-1,[H+]=[C6H7O6-]at equilibrium, so we can rewrite this as:     Ka =[H+]2/[HA] We are given that when 10 mL of base has been added, the pH of thesolution is 3.72. First we need to calculate the moles of ascorbicacid remaining after addition of KOH, as you would with a normaltitration question:     n(KOH) = 0.010 L * 0.1103 mol/L =1.103x10-3 mol     n(acid reacted) = n(KOH) = 1.103x10-3mol     n(acid remaining) = n(acid originally) - n(acidreacted) = 3.135x10-3 - 1.103x10-3 =2.032x10-3 mol We are given that the pH of the solution is 3.72, so we cancalculate the concentration of H+ (on the right-handside of the equilibrium expression):     pH = -log[H+] [H+] = 10-pH = 10-3.72M In a volume of 0.030 L, we can also calculate the moles ofH+, and use this to calculate the equilibriumconcentration of ascorbic acid:     n(H+) = 10-3.72 * 0.030 = 5.716x10-6mol = n(acid dissociated)     n(acid) = n(acid remaining) - n(aciddissociated) = 2.032x10-3 - 5.716x10-6 = 2.026x10-3     [acid] = 2.026x10-3 mol / 0.030 L = 0.0675 Now we can calculate the acid dissociation constant:     Ka = [H+]2 /[HA] = (10-3.72)2 / 0.0675 =5.37x10-7 Hrm. I checked this on the net and it doesn't seem to agree withthe literature value (7.6x10-5); not sure why -sorry!

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