Question

Data set used for this experiment.	1
Concentration of HCl (M)	0.445
Volume of glass cleaner (mL)	25.00

	Volume (mL)	pH
Data point 1 (0.00 mL)	0.00	11.26
Data point 2	2.00	10.73
Data point 3	4.00	10.41
Data point 4	6.00	10.20
Data point 5	8.00	10.07
Data point 6	10.00	9.93
Data point 7	12.00	9.82
Data point 8	14.00	9.71
Data point 9	16.00	9.61
Data point 10	18.00	9.48
Data point 11	20.00	9.38
Data point 12	22.00	9.21
Data point 13	24.00	9.04
Data point 14	26.00	8.84
Data point 15	26.50	8.77
Data point 16	27.00	8.65
Data point 17	27.50	8.60
Data point 18	28.00	8.50
Data point 19	28.50	8.39
Data point 20	29.00	8.20
Data point 21	29.50	7.92
Data point 22	30.00	6.25
Data point 23	30.50	3.81
Data point 24	31.00	3.32
Data point 25	31.50	3.22
Data point 26	32.00	3.13
Data point 27	32.50	3.07

1. Equivalence point (mL) ________

2. Calculate the concentration of the weak base. ________

3. Volume of HCl needed to reach the half of the equivalence point. _______

4. Determine the K_b of the weak base. ________

Answer 1

Answer

1. Equivalence point (mL) 30 mL

2. Calculate the concentration of the weak base. 0.534 M

3. Volume of HCl needed to reach the half of the equivalence point.15 mL

4. Determine the K_b of the weak base.Kb = 1,00,000.

Explanation

Primary graph

12 pH=pkdo =9.0 00 6 4 2 0 Half eq. = 15 ml 15 20 Eq.point 30 mL 30 0 LO 5 25 35 Volume of HCI

Derivative graph

6 5 4 Delta pH/Delta V 2 1 0 0 5 10 15 20 25 30 35 Volume of HCI

V	pH	$\Delta V$	$\Delta pH$	ΔpH/ΔΥ
0	11.26	2	0.53	0.265
2	10.73	2	0.32	0.16
4	10.41	2	0.21	0.105
6	10.2	2	0.13	0.065
8	10.07	2	0.14	0.07
10	9.93	2	0.11	0.055
12	9.82	2	0.11	0.055
14	9.71	2	0.10	0.05
16	9.61	2	0.13	0.065
18	9.48	2	0.1	0.05
20	9.38	2	0.17	0.085
22	9.21	2	0.17	0.085
24	9.04	2	0.2	0.1
26	8.84	0.5	0.07	0.035
26.5	8.77	0.5	0.12	0.24
27	8.65	0.5	0.05	0.1
27.5	8.6	0.5	0.1	0.2
28	8.5	0.5	0.11	0.22
28.5	8.39	0.5	0.19	0.38
29	8.2	0.5	0.28	0.56
29.5	7.92	0.5	1.67	3.34
30	6.25	0.5	2.44	4.88
30.5	3.81	0.5	0.49	0.98
31	3.32	0.5	0.1	0.2
31.5	3.22	0.5	0.09	0.18
32	3.13	0.5	0.05	0.1
32.5	3.07

Calculation for concentration of weak acid

[HCl] = 0.445 M

volume of HCl = 30.0 mL (at eq. point)

volume of weak acid = 25.0 mL

Concentration of weak acid = (0.445*30.0)/25

                                        = 0.534 M

Calculation for kb

pka = 9.0

we know pka + pkb = 14

               9.0 + pkb= 14

                      pkb = 14 - 9.0

                      pkb = 5.0

pkb = -log[ka]

5.0 = -logka

ka = 1,00,000