Polyproticacids What are the majorf species in solution during the titration of
ID: 958040 • Letter: P
Question
Polyproticacids What are the majorf species in solution during the titration of citric acid with a strong base? Citric acid is a weak orgonic acid that occurs naturally In citrus fruits and is mainly used as a flavoring and preservative in food and beverages. especially soft drinks. Citric acid is a triprotic acid. Although its formula C_6H_8O_7 Indicates of 8 hydrogens, only the hydrogens attached to the three carboxylic groups (COOH) are acidic. In this part of the lab. you will calculate 8 data points of a titration of 100 ml of 0.50 M citric acid and 0.50 M NaOH and then draw the graph. Acid dissociation constants of citric acid. Acid dissociation: Start by writing the equations for the three steps of dissociation of citric acid. Make sure you give each species the correct charge. Initial pH. Calculate the initial pH of the 0.50 M citric acid solution. The Initial pH is determined by K and the initial concentration of the acid. Midpoints: Determine the pH of the 3 midpoints of the titration. At each midpoint, the pH = pK, of the species you are titrating. Equivalence points: Calculate the pH of the 3 equivalence points of the titration. At the first and the second equivalence point, the pH is the average of the PK_a values above and below. At the third equivalence point (the end point), the pH is determined by the K_b of the conjugate base of the weakest acid and its concentration. Past the equivalence point: Calculate the pH after 400 mL of NaOH were added. At this point the pH depends only on the concentration of excess NaOH. Titration curve: The titration curve is a pH vs. volume of 0.50 M NaOH graph. Plot the 8 points and connect them to determine the titration curve of citric acid. The curve should be relatively flat around each of the midpoints (buffering region) and should sharply increase around the equivalence points. For each of the 8 points write the major species (other than H_20 and Na^+) next to the point in the graphExplanation / Answer
Given Ka1 = 7.5x10-4
=> pKa1 = - logKa1 = - log(7.5x10-4) = 3.125
Ka2 = 1.7x10-5
=> pKa2 = - logKa2 = - log(1.7x10-5) = 4.77
Ka3 = 4.0x10-7
=> pKa3 = - logKa3 = - log(4.0x10-7) = 6.40
4: pH at first equivalence point = (Ka1 + Ka2) / 2 = (3.125+4.77) /2 = 3.947 (answer)
pH at second equivalence point = (Ka2 + Ka3) / 2 = (4.77 + 6.40) /2 = 5.585 (answer)
During the 3rd equivalence point all of the C6H8O7 is converted to salt C6H5O73-(aq).
Since citric acid is a triprotic acid and both citric acid and NaOH have same concentration, the total volume of NaOH required to reach 3rd equivalence point is three times the volume of citric acid.
Hence total volume of the solution after 3rd equivalence point, Vt = 3x100 mL(NaOH) + 100 mL(citric acid)
= 400 mL = 0.400 L
Total moles of C6H5O73-(aq) = total moles of citric acid = MxV = 0.5 mol/L x 0.100 L = 0.05 mol
Hence after 3rd equivalence point [C6H5O73-(aq)] = 0.05 mol / 0.400 L = 0.125 M
Now C6H5O73-(aq) undergoes hydrolysis and the pH can be calculated from salt hydrolysis equation.
pH = (1/2)x(pKw + pKa3 + log[C6H5O73-(aq)]
=> pH = (1/2)x(14 + 6.40 + log0.125M) = 9.75
Hence pH after 3rd equivalence point = 9.75 (answer)
5: Post the equivalence point, the volume of 0.5M NaOH added = 400 mL = 0.400 L
Hence moles of NaOH added = MxV = 0.5 mol/L x 0.400 L = 0.2 mol
Volume of 0.5M NaOH added to 100 mL of 0.5 M citric acid to reach equivalence point = 3x100 mL
= 300 mL = 0.3 L
Hence total volume of the solution = 100 mL citric acid + 300 mL NaOH (till equivalencepoint) + 400 mL (after equivalence point)
= 800 mL = 0.800 L
Hence concentration of NaOH = [OH-] = moles of NaOH / volume = 0.2 mol / 0.800L = 0.25 M
=> pOH = - log[OH-] = - log(0.25M) = 0.602
=> pH = 14 - pOH = 14 - 0.602 = 13.40 (answer)