Question

nitial data

Enter the concentration of NaOH used in the titration. (M)

0.5

Enter the mass of unknown acid you used in grams. (g)

0.75

pH vs Volume Measurements

Table 1. pH readings

	Volume (mL)	pH
Initial measurement	0	1.43
Data point 1	0.99	1.23
Data point 2	2	1.41
Data point 3	2.97	1.5
Data point 4	3.99	1.59
Data point 5	4.99	1.69
Data point 6	5.99	1.79
Data point 7	6.98	1.89
Data point 8	7.99	2
Data point 9	9.03	2.14
Data point 10	9.97	2.29
Data point 11	10.97	2.5
Data point 12	11.98	2.85
Data point 13	13.03	4.32
Data point 14	14.02	5.35
Data point 15	15.01	5.87
Data point 16	15.99	6.08
Data point 17	16.98	6.25
Data point 18	17.97	6.4
Data point 19	18.97	6.53
Data point 20	20.02	6.67
Data point 21	20.97	6.81
Data point 22	21.98	6.96
Data point 23	22.99	7.13
Data point 24	24.03	7.36
Data point 25	25.02	7.74
Data point 26	26.01	10.64
Data point 27	27	11.49
Data point 28	28.03	11.75
Data point 29	29	11.91
Data point 30	30.02	12.02
Data point 31	30.99	12.1
Data point 32	32.01	12.17
Data point 33	33.01	12.22
Data point 34	34.02	12.27
Data point 35

(13pts) Part A - Interpreting the titration curve

Use the titration curve to determine the equivalence points for the unknown acid, the volumes at 50% of the equivalence points, K_a1, and K_a2 for the unknown acid. Because we recorded the pH at about 1 mL increments, we will need to do some estimating of pH values. When doing the estimating, a good approach is to average the known values on either side of the point you want to estimate. There will be some error in our readings, but it should be small enough for our purposes.

(1pts)

Select the unknown you used for this experiment.

Choose...ABC

(2pts)

Identify the second equivalence point on the titration curve. Use the two volume measurements where the largest change in pH occurs. Take the average of these two volumes to find where the equivalence occurs. What is the volume needed to reach the second equivalence point?

(2pts)

Identify the first equivalence point on the titration curve. The first equivalence point should occur at half the volume needed for the second equivalence point. Look for the largest pH change near this value, and take the average of the volumes above and below it. What is the volume of NaOH needed to reach the first equivalence point?

(1pts)

Volume at 50% of the first equivalence point.

(1pts)

Estimate the pH at 50% volume of the first equivalence point. This represents pK_a1.

(2pts)

What is the K_a1?

(1pts)

Volume at 50% between the first and second equivalence point.

(1pts)

Estimate the pH at the volume 50% between the first and second equivalence point. This represents pK_a2.

(2pts)

What is the K_a2?

(5pts) Part B - Molar mass of unknown acid

Using the volume of the second equivalence point, find the moles of acid present, and the molar mass of the unknown acid.

(1pts)

Moles NaOH needed to reach the second equivalence point.

(2pts)

Moles of unknown acid.

(2pts)

Molar mass of unknown acid.

(5pts) Part C - Identity of unknown acid

Use the following data to identify your unknown acid.

Acid	Formula	Ka1	Ka2
Ethylglutamic acid	C7H13NO4	1.43 × 10−04	1.45 × 10−08
Maleic acid	C4H4O4	1.50 × 10−02	2.60 × 10−07
Malonic acid	C3H4O4	1.50 × 10−03	2.00 × 10−06
Oxalic acid	C2H2O4	5.90 × 10−02	6.40 × 10−05
Tartaric acid	C4H6O6	1.00 × 10−03	4.60 × 10−05

(5pts)

What is most likely the identity of your unknown acid?

Answer 1

From the data the second equivalence point is between 26.01 and 25.02 mL

                              Therefore the average of these two = 25.52 mL

          The first equivalence point lies between 11.98 and 13.03 mL

                            Therefore the average of these two = 12.51 mL

50% of the volume of first equivalence point ( first half equivalence point ) = 12.51 /2 = 6.255 mL

pH at this point = 1.79 + [1.89 - 1.79 ] [ 6.255 - 5.99] / [6.98 - 5.99]    [ by interpolation ]

                        = 1.79 + [ 0.1 x 0.265]/0.99

                       = 1.79 + 0.027

                      = 1.82

Therefore pKa1 of the unknown acid = 1.82

Ka1 = 10^-pKa1 = 10^-1.82 = 1.51 x 10^-2

50% of the volume between first and second equivalence point is = [25.52 + 12.51] /2 = 19.015 mL

That is the volume at second half equivalence point = 19.015 mL

pH at second half equivalence point = 6.53 + [ 19.015 - 18.97] [ 6.67-6.53] / [20.02 - 18.97]

                                                   = 6.53 + [0.045 x 0.14] / 1.05

                                                   = 6.53 + 0.006

                                                     = 6.536 = 6.54

Thus the pH at second equivalence point = 6.54

Therefore pKa2 of the unknown acid = 6.54

Thus Ka2 = 10^-pKa2 = 10^-6.54 = 2.88 x 10^-7

Consider the equation for a diprotic acid with NaOH

H2A + 2 NaOH = Na2A + 2H2O

Thus one mole of the unknown acid consumes 2 moles of NaOH

Therefore moles of unknown acid = [1/2] moles of NaOH used

     The volume of 0.5 M, NaOH used to reach the second equivalence point is 25.52 mL

therefore moles of NaOH used = 25.52 mL x 0.5 mol / 1000 mL

                                             = 0.01276 moles

Therefore moles of unknown acid = [1/2] x 0.01276 moles

                                                = 0.00638 moles

Given mass of the unknown acid = 0.75 g

Therefore molar mass of the unknown acid = g / mol = mass / mole = 0.75 / 0.00638 = 117.55 g / mol

Thus the molar mass of the unknown acid = 117.55 g / mol.

The molar mass of maleic acid from the given formula C4H4O4 is 48 + 4 + 64 = 116 g / mol

By comparing the Ka1, Ka2, molar mass of the unknown acid, the given unknown must be maleic acid.

Thus the most likely unknown acid is Maleic acid