Question

Consider the titration of 40.0 mL of 0.523 M ascorbic acid (H2C6H6O6) with 0.885 M potassium hydroxide (KOH): pKa1 = 4.10 pKa2 = 11.80

What is the solution pH before the titration begins?

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What is the pH at the first half-equivalence point? What is the pH at the first equivalence point?

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What is the pH at the second half-equivalence point?

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What is the pH at the second equivalence point? _________________

What is the pH when 70.0 mL of strong base has been added?

Answer 1

Consider the titration of 40.0 mL of 0.523 M ascorbic acid (H2C6H6O6) with 0.885 M potassium hydroxide (KOH): pKa1 = 4.10 pKa2 = 11.80

What is the solution pH before the titration begins?

using general teartment for such weak acid :

[H+]= (Ka1*[Acid])^1/2

Ka1 = 10^-pka1 = 10^-4.10 = 7.94*10^-5

[H+] = ( 7.94*10^-5*[0.523 ])^1/2 = 6.45*10^-3 M

pH = - log [H+ ] = 2.19

What is the pH at the first half-equivalence point? What is the pH at the first equivalence point?

pH at the first half-equivalence point = pKa1 = 4.10

pH at the first equivalence point = (pKa1+pKa2)/2 = 7.95

What is the pH at the second half-equivalence point?

pH at the first half-equivalence point = pKa2 = 11.80

What is the pH at the second equivalence point?

now , C6H6O6^2- is main active species :

Veq 2 (KOH) = 2* 40.0 mL *0.523 M /0.885 M = 47.3 mL

[ C6H6O6^2- ] =   40.0 mL *0.523 M/ 87.3 mL = 0.240 M

Kb = 10^-14 / Ka2 = 6.31*10^-3

[OH-] = ( Kb*[ C6H6O6^2- ])^1/2 = 0.0389 M

pH = 14-pOH = 12.6

What is the pH when 70.0 mL of strong base has been added?

[OH-] = (70.0 - 47.3 mL) * 0.885 M / 110 mL = 0.183 M

pH = 13.3