Strong Acid Strong Base Lab PDF

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Strong Acid/ Strong Base Lab Report...

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Chemical Equilibrium: Iron (III) with SCN Lab Report Introduction: In this lab, a strong acid (HCl) of unknown concentration was titrated into a strong base (NaOH) of known concentration, in order to determine the molarity of HCl. Titration is a common laboratory technique to determine the concentration of an acid/base by neutralizing the titrand. This can be done by either determining the equivalence point with an indicator (in our case, phenolphthalein) or by measuring the pH of the solution while titrating. At the equivalence point, the most rapid change in pH will occur. In this experiment, we will be neutralizing HCl with NaOH. Neutralization of the solution occurs as a consequence of reacting acids and bases. This can be conceptualized in part by the Arrhenius theory of acids and bases, wherein acids form the hydronium ion in aqueous solution and bases form hydroxide, which react together to give water. More generally, neutralization reactions between an acid and base form a salt and water. The reaction we will be observing is as follows: HCl + NaOH NaCl + H2O Experimental: There was one deviation made from the standard procedure listed in the lab manual. For the titration with a pH probe, Data Analysis: Titration Without a pH Probe Figure 1: Volume of NaOH Required to Reach Equivalence Point TRIAL

Vi in Buret (mL)

Vf in Buret (mL)

ΔV = NaOH Titrated (mL)

1

0.60

23.90

23.30

2

0.24

23.38

23.14

3

0.02

23.12

23.10

Mean

0.29

23.47

23.18

Using the average amount of NaOH needed, we can calculate the initial concentration of HCl with the equation for equivalence points between acids/bases:

(Vbase)(Mbase) = (Vacid)(Macid) For our experiment, we used NaOH of concentration .22M and needed an average of 23.18 mL. Thus, the equation yields: (23.18 mL)(.22 M) = (X)(100 mL) X = .051M = [HCl Solution] This indicates that the HCl solution in the beaker had a molarity of .051M, however, since this solution had diluted the original HCl solution with 3 parts water to 1 part HCl, we need to multiply this concentration by 4 to obtain the concentration of the original HCl solution. .051*4 = .204M = [HCl] Titration with a pH Probe Following the titration with only phenolphthalein indicator, we then conducted the same titration with a pH probe. The following data was obtained for the titration of HCl and NaOH with a pH probe: Figure 2: pH of Solution vs Amount of NaOH added; Trial One mL in buret

pH

ΔpH

ΔV (mL)

ΔpH/ΔV

12.78

1.57

29.9

2

0.43

17.12

0.02511682243

32.42

2.38

0.38

2.52

0.1507936508

32.82

2.5

0.12

0.4

0.3

33.52

2.8

0.3

0.7

0.4285714286

34.06

3.23

0.43

0.54

0.7962962963

34.2

3.6

0.37

0.14

2.642857143

34.32

3.99

0.39

0.12

3.25

34.36

4.89

0.9

0.04

22.5

34.4

5.71

0.82

0.04

20.5

34.46

6.32

0.61

0.06

10.16666667

34.52

6.6

0.28

0.06

4.666666667

34.58

6.87

0.27

0.06

4.5

34.64

7.33

0.46

0.06

7.666666667

34.7

7.52

0.19

0.06

3.166666667

34.76

7.64

0.12

0.06

2

34.84

7.81

0.17

0.08

2.125

35.12

8.16

0.35

0.28

1.25

35.28

8.51

0.35

0.16

2.1875

35.4

9.52

1.01

0.12

8.416666667

35.62

10.14

0.62

0.22

2.818181818

35.72

10.4

0.26

0.1

2.6

36.94

10.69

0.29

1.22

0.237704918

37.02

10.98

0.29

0.08

3.625

38.72

11.46

0.48

1.7

0.2823529412

The data from the first two columns of the above table was used to obtain the following graph:

Given that the equivalence point is where the most rapid changes in pH occur, we can see that the equivalence point is somewhere around 34 mL of NaOH. We can confirm this with the graph of the change in pH per change in volume of NaOH versus volume of NaOH:

From this graph, we see that the most rapid change in pH occurs at a volume (of NaOH) of 34.36 mL. Since the original volume of NaOH in the buret read 12.78 mL, we must subtract this value from our obtained value. 34.36 mL - 12.78 mL = 21.58 mL This is the amount of NaOH needed to reach the equivalence point. We must repeat such analysis for the second trial of our titration with a pH probe: Figure 3: pH of Solution vs Amount of NaOH added; Trial Two mL in Buret

pH

ΔpH (mL)

ΔV

ΔpH/ΔV

0.08

1.51

13.58

1.8

0.43

13.5

0.03185185185

17.48

2.11

0.31

3.9

0.07948717949

19.5

2.32

0.21

2.02

0.103960396

20.66

2.6

0.28

1.16

0.2413793103

21.32

2.91

0.31

0.66

0.4696969697

21.62

3.19

0.28

0.3

0.9333333333

21.74

3.5

0.31

0.12

2.583333333

21.86

4

0.5

0.12

4.166666667

21.98

5.37

1.37

0.12

11.41666667

22.02

5.8

0.43

0.04

10.75

22.1

6.09

0.29

0.08

3.625

22.18

6.81

0.72

0.08

9

22.24

7.4

0.59

0.06

9.833333333

22.3

8

0.6

0.06

10

22.36

8.9

0.9

0.06

15

22.42

9.34

0.44

0.06

7.333333333

22.48

9.6

0.26

0.06

4.333333333

22.58

9.95

0.35

0.1

3.5

22.66

10.17

0.22

0.08

2.75

22.82

10.45

0.28

0.16

1.75

23

10.6

0.15

0.18

0.8333333333

23.12

10.72

0.12

0.12

1

23.42

10.92

0.2

0.3

0.6666666667

24.02

11.14

0.22

0.6

0.3666666667

25.4

11.41

0.27

1.38

0.1956521739

Similar graphs to those above were constructed for the data from trial two:

From these graphs, we see that the equivalence point-- using the same method as done for trial one-- is at 22.36 mL. Since the initial volume of NaOH in the beaker was 0.08 mL, this means 22.28 mL was required to reach equivalence. We can now average our values from trials one and two to get the average amount of NaOH needed to reach equivalence: (21.58+22.36)/2 = 21.97 mL, the average amount of NaOH needed to reach equivalence. Using the same equation as we did to calculate the initial concentration of HCl in the first half of this lab, we will now calculate the concentration of HCl solution: (Vbase)(Mbase) = (Vacid)(Macid) (21.97 mL)(.22M) = (100 mL)(X) X = .048 M = concentration of HCl solution Since we similarly diluted the HCl in this part, we must multiply the calculated concentration by 4. 4*(.048) = .192M = [HCl]

Discussion: From our titration of HCl and NaOH without a pH probe and just the phenolphthalein indicator, we obtained a value for the initial concentration of HCl to be .204M. When we conducted the same titration with a pH probe, we obtained a value for the initial concentration of HCl to be .192M. This value had a percent difference from the first value of 5.88%. This suggests that both methods yielded relatively consistent values for the initial concentration of HCl. This makes sense because the same HCl solution was used for both titrations. 1. As mentioned in the prior discussion of the titration without the pH meter, phenolphthalein was used as an indicator so as to determine the endpoint. The choice of indicator is an important factor in titrations such as that which was just described. The most important factor in choosing an indicator is the pH region in which it changes color. As described in this report, the indicator is used to visually demonstrate the endpoint of a titration, and so an indicator should be used which changes over the range of the pH change of the titration. Like the phenolphthalein used herein, the range in which many indicators change color is not at about the neutral pH of 7. Phenolphthalein, however, was a useful indicator in this titration because of the steepness of the titration curve about the equivalence point, which makes it so that increase of pH to the point where phenolphthalein changes involves a negligible increase in NaOH volume. An indicator with a similar range (9-10, for example) would also be useful for a titration of a weak acid with a strong base. This is because such a titration would have an equivalence point lying around 9 or 10, which would make an indicator like phenolphthalein an ideal choice for determining the end of this titration. The equivalence point is above 7 because the conjugate base of a weak acid is a relatively strong base. 2. We are using pH = 7 for the equivalence point of HCl vs NaOH titration because HCl is a strong acid and NaOH is a strong base. This means both will dissociate completely in solution and the the H+ and OH- ions will completely neutralize each other to form water and an NaCl salt, which will yield a pH of 7 (or the pH of pure water). The possible sources of error in this lab stem from the fact that the titration without a pH probe was a qualitative analysis of the pH of the solution. We determined the equivalence point to be the point where the solution turned light pink, but this is a generally broad point, and we could've passed the true equivalence point. Thus, the value for the first titration was slightly higher....