Question

1) A student titrated 20.0 mL of 0.410 M HCl with 0.320 M NaOH and collected the following data:

Number	V of NaOH solution added, mL	PH	#	V of NaOH added, mL	PH
1	0.00	.39	12	22.00	1.56
2	2.00	.46	13	24.00	1.93
3	4.00	.54	14	24.50	2.09
4	6.00	.62	15	25.00	2.35
5	8.00	.70	16	25.50	3.06
6	10.00	.78	17	26.00	11.40
7	12.00	.87	18	26.50	11.80
8	14.00	.96	19	27.00	12.00
9	16.00	1.07	20	28.00	12.20
10	18.00	1.19	21	29.00	12.30
11	20.00	1.35	22

a) Prepare a titration curve for this titration using the data provided and excel program. Correctly label the two axes.

b) From the above titration curve, identify the volume of titrant required to reach the equivalence point of the titration. Now use the information given in the problem and calculate the volume of titrant needed to reach the equivalence point. compare the two values. which one is more accurate?

c) calculate the pH valuse when 21.50 mL of titrant has been added?

d) what should you do with the NaOH solution remaining in your burette after you have completed your titrations?

Answer 1

a) The curve is observed here:

b) It is observed in the titration curve, the point where there is a sudden change in pH: a V NaOH = 26 mL

It can also be calculated by the dilution ratio:

V NaOH = Ca * Va / Cb = 0.41 M * 20 mL / 0.32 M = 25,625 mL

more accurate is through calculation.

c) First the initial moles of acid and the added base volume are calculated:

n HCl = 0.41 M * 0.02 L = 0.0082 mol

n NaOH = 0.32 M * 0.0215 L = 0.0069 mol

the final moles of HCl are calculated:

n End HCl = 0.0082 - 0.0069 = 0.0013 moles

the concentration to the final volume of 41.5 mL:

[HCl] = 0.0013 mol / 0.0415 L = 0.0313 M

and the pH = - Log (0.0313) = 1.50

d) The rest of the titrant solution can be emptied in a clean and dry vial, covered, labeled and reserved.