Question

hi im having trouble with a titration lab. van somebody answer snd explain how to do question 1. THE INITIAL CONCENTRATION OF ACID IS .5M.

hydrogen attached to the three carboxylic group hydrogens you will calculate 8 data poin titration of NaOH and then draw the grap Table 15.3 Acid dissociation constants of citric 7.5 x 10-4 1.7 x 10-5 4.0 x 107 1. Acid dissociation: Start by writing the eq citric acid Make sure you give each speci

THE QUESTION TO DIRECTLY ANSWER IS QUESTION NUMBER 2.

Answer 1

Titration of citric acid

1. Dissociation equations,

let H3Cit be the citric acid neutral form

First dissociation : H3Cit <==> H+ + H2Cit-

Second dissociation : H2Cit- <==> H+ + HCit^2-

Third dissociation : HCit^2- <==> Cit^3- + H+

2. pH of 0.5 M Citric acid

Ka1 = [H+][H2Cit-]/[H3Cit]

7.5 x 10^-4 = x^2/0.5

x = [H+] = 0.0194 M

pH = -log[H+] = 1.713

3. pH at midpoints

First midpoint = pKa1 = -log[Ka1] = 3.125

Second midpoint = pKa2 = -log[Ka2] = 4.770

Third midpoint = pKa3 = -log[Ka3] = 6.400

4. pH at equivalence points

First equivalence point pH = (pKa1 + pKa2)/2 = 3.950

Second equivalence point = (pKa2 + pKa3)/2 = 5.585

Third equivalence point

Kb1 = Kw/Ka3 = 1 x 10^-14/4 x 10^-7 = x^2/0.5

x = [OH-] = 1.12 x 10^-4 M

pOH = -log[OH-] = 3.95

pH = 14 - pOH = 10.05

5. Past equivalence point,

pH is dependent upon concentration of excess NaOH only in solution.