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Malonic acid can ionise in two stages as it is a dibasic acid.
The values of pKa1 and pKa2 of malonic acid are 2.85 and 5.70 respectively. Calculate the pH of the first and second equivalence points of the titration of 10cm^3 of malonic acid of concentration 0.1 M with sodium hydroxide of concentration 0.1 M.
my attempt at a solution:
maximum buffering capacity before first endpoint: pH=pKa1 = 2.85
[OH-] = [(Kw/Ka1)*(0.05)]^(0.5)
therefore pOH = 6.23
pH = 14-6.23 = 7.77
im stuck here as the answer given for the pH of the first endpoint is 3.50 and I have no idea how to work it out
The values of pKa1 and pKa2 of malonic acid are 2.85 and 5.70 respectively. Calculate the pH of the first and second equivalence points of the titration of 10cm^3 of malonic acid of concentration 0.1 M with sodium hydroxide of concentration 0.1 M.
my attempt at a solution:
maximum buffering capacity before first endpoint: pH=pKa1 = 2.85
[OH-] = [(Kw/Ka1)*(0.05)]^(0.5)
therefore pOH = 6.23
pH = 14-6.23 = 7.77
im stuck here as the answer given for the pH of the first endpoint is 3.50 and I have no idea how to work it out