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Solubility Product Lab Report...

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Solubility Product of Calcium Benzoate Lab Report Introduction: The purpose of this experiment is to determine the solubility product of a calcium benzoate, or Ca(C6H5COO)2. For the purposes of this report, we will abbreviate the benzoate ion as Bz- ; thus, we will call calcium benzoate CaBz2. The principles of this experiment pertain to solubility. A slightly soluble salt, such as CaBz2 , will dissociate in solution by the given formula: CaBz2(s) --> Ca2+ (aq) + 2Bz-(aq) This dissociation should proceed until equilibrium is reached, or when Ksp (the solubility product) equals the following quantity: Ksp = [Ca2+][Bz- ]2 If the quantity [Ca2+  ][Bz- ]2 is greater than Ksp, some ions will combine to form precipitate CaBz2, until equilibrium is reached. Conversely, if the same quantity is less than Ksp, more CaBz2 molecules will dissociate to increase the quantity of ions to re-establish equilibrium. In this experiment, we will conduct six titrations of three different initial concentrations of Ca2+  - and Bz ions. We will be titrating these solutions against a solution of EDTA of known concentration. EDTA is a metal chelator, meaning it forms complexes with metals such as Ca2+  . These complexes consist of a central metal ion surrounded by chelator ligands. The metal ion acts as a Lewis acid and accepts electron pairs during titration; the ligands act as Lewis bases and donate electron pairs during titration. We will know when the equilibrium is reached by using a Calmagite indicator. Calmagite (H3T) will first form a red solution, due to the following equation: Ca2+ + HT2-  CaT- + H+ The CaT- complex will appear red in solution. The titration of EDTA (Na2H2E) against Ca2+  will cause the following reaction: Ca2+  + H2E2-  CaE2-  + 2H+ Once the EDTA has chelated all of the free Ca2+  ions, the Calmagite will break up the Ca2+-EDTA complexes and form a new HT2-  complex. This happens because the affinity between EDTA and Ca2+  is less than that of EDTA and Calmagite. At equivalence, the following equilibrium will exist: CaT- + H2E2-  CaE2-  + HT2-  + H+

The red CaT- complex will dissociate and a new HT2-  complex will form, that is blue. This is when we will know that equilibrium has been obtained. In addition, we will also know that the number of moles of Ca2+  equals the number of moles of EDTA at this point because all of the Ca2+ ions need to be complexed with EDTA to reach equivalence. This is crucial to our  . calculation of [Ca2+  ]f, [Bz- ]f, and our Ksp One last concept must be noted. We will be adding a few crystals of MgCl2 to our titration because the Mg-EDTA complex forms a deep blue color that will better indicate that we have reached equivalence. The addition of Mg2+  will not affect our titration because the Mg-EDTA  ions have been chelated by EDTA. complex only forms after  all of the Ca2+ Experimental: In this experiment, we must first prepare our titrant, a solution of EDTA of known concentration. To prepare the EDTA, simply add a massed amount of EDTA salt to a volumetric flask and dissolve into 250 mL of water. We will titrate three different solutions of CaBz2 of known initial concentrations of Ca2+  and Bz- . After filtering these three solutions to remove any precipitate, we will put a 20-mL aliquot into our titration beaker, along with a few crystals of MgCl2 to sharpen our endpoints. 50-mL of EDTA will be placed into the titration buret and titrated until a bluish-purple color is obtained. Refill the buret and repeat the procedure for both trials of all three solutions. Record the amount of EDTA needed to titrate each solution, along with visible observations. (No deviations were made from the UChicago General Chemistry Lab Manual, "General Chemistry Experiments: 1st Edition" (Zhao, Dragisich). Results: Mass of EDTA: 9.00 g Volume of EDTA Solution: 250 mL Figure 1: Data for 20-mL Calcium Benzoate Solution Aliquots Titrated Against EDTA Solution Solution #

T (°C)

[Ca2+  ]i

[Bz- ]i

Vi (mL)

Vf (mL)

∆V (mL)

1

21.9

0.20

0.20

2.2

43.0

40.8

1

21.9

0.20

0.20

0.0

40.1

40.1

2

21.9

0.07

0.32

0.00

12.3

12.3

2

21.9

0.07

0.32

12.3

24.4

12.1

3

21.9

0.125

0.250

25.0

48.8

23.8

3

21.9

0.125

0.250

0.1

23.0

22.9

Data Analysis: We must first calculate the concentration of our EDTA solution, in order to be able to make titration calculations. The concentration of the EDTA solution can be calculated as follows: (9.00 g) (1 mol EDTA / 372.24 g) = 0.0242 mol EDTA (0.0242 mol EDTA / 0.250 L) = 0.0967 M EDTA solution In order to calculate the Ksp values for our data, we must use the following equation: Ksp = [Ca2+]f[Bz- ]f 2 We know that the number of moles of Ca2+  at equilibrium is equal to the number of moles of EDTA needed to reach the equilibrium. We know this because the solution changes to blue (reaches equilibrium) when EDTA has no more Ca2+  to complex; if all of the Ca2+  is complexed with EDTA, the two will have equal concentrations at equilibrium. Using this information, a sample calculation for Solution 1, trial 1 is below: ∆V = 40.8 mL [EDTA] = 0.0967 M moles of EDTA = (0.0964 mol / 1 liter) (.0408 L) = 0.00393 = moles of Ca2+  at equilibrium 2+ 2+ (0.00393 moles / .020 L) = 0.197 M = [Ca ]f, or the concentration of Ca at equilibrium Using this concentration, we can determine the concentration of Bz- at equilibrium for Solution 1 using a simple ICE table. CaBz2(s)

-->

Ca2+ (aq)

2Bz- (aq)

Initial

--

0.20 M

0.20 M

Change

--

∆C

2∆C

Equilibrium

--

0.197 M

?

∆C = 0.197 M - 0.20 M = -0.0033 M 2∆C = -0.0067 M [Bz-]f = 0.20 + 2∆C = 0.193 M We know that the Ksp value is given by the quantity [Ca2+  ]f[Bz- ]f 2. We can now calculate the Ksp for Solution 1, Trial 1 because we have the equilibrium concentrations of Ca2+  and Bz- . Ksp = [Ca2+]f[Bz-]f2 = (0.197)(.193)2 = 0.00221 = 7.37 x 10-3

The above calculations were repeated for both trials of each solution and summarized into the table below: Figure 2: Solubility Product Ksp of CaBz2 Based On the Values of [Ca2+  ]f and [Bz- ]f in Solution Solution #

[Ca2+  ]i

[Bz- ]i

[Ca2+  ]f

[Bz- ]f

Ksp

1

0.20

0.20

0.197

0.193

7.37E-3

1

0.20

0.20

0.193

0.187

6.73E-3

2

0.07

0.32

0.593

0.299

5.3E-3

2

0.07

0.32

0.583

0.297

5.1E-3

3

0.125

0.250

0.115

0.229

6.0E-3

3

0.125

0.250

0.110

0.221

5.4E-3

Average Ksp = 5.99E-3 We must standardize our average Ksp value to 20°C to be able to compare our value to a known value for the Ksp of calcium benzoate. Assuming the Ksp increases by 2.5% for each degree (Kelvin/Celsius), we can calculate our new Ksp by the following equation: Ksp (T2) = (Ksp (T1))(1.025)∆T  , where ∆T = T2 - T1, T2 is equal to 20°C, and T1 is equal to our average measured temperature. Figure 3: Temperatures Measured for Each Solution Solution 1

20.7°C

Solution 2

21.2°C

Solution 3

21.3°C

Average

21.1°C

Ksp (20°C) = Ksp (21.1°C)(1.025)-1.9  -1.9 (5.99E-3)(1.025) Ksp (20°C) =  Ksp (20°C) = 5.83E-3 Discussion: The purpose of this experiment was to determine the solubility product of the dissociation of calcium benzoate. We ran six trials of 3 different initial concentrations of Ca2+  and Bz- ions. Using the final concentration of EDTA in solution, we were able to calculate the amount of Ca2+  in solution at equivalence. Given that we knew the initial-- and now the final-concentrations of Ca2+  in solution, we were able to use the ICE table to calculate the final concentration of Bz- in solution. Using the final concentrations of both Ca2+  and Bz- , we were 2+ - 2 able to calculate Ksp using the quantity [Ca ][Bz ] . Our average Ksp among all six trials was

5.99E-3. All of our solutions had their solubility products calculated at 21.1 Celsius. We know that solubility increases with temperature, therefore, the solubility at 20.0 degrees Celsius would be lower than our average. After standardizing, our value for Ksp (20°C) was 5.83E-3. Per my research, the accepted value for the solubility product of calcium benzoate is 2.12E-3 [2]. The percent error can be calculated as follows: [|Observed-Accepted|/Accepted]*100 [(5.83E-3-2.12E-3)/2.12E-3]*100 = 175% error Some sources of error for our experiment must have arisen from the qualitative nature of our titration. We were aiming for a bluish-purple solution during titration, however, it is entirely possible we repeatedly over-titrated. Had we over titrated, we would over-approximate the concentration of Ca2+  at equilibrium. This would give us a greater ∆C and a greater 2∆C in our ICE tables, therefore, overestimating the concentration of Bz- at equilibrium. Since this quantity is squared, we can estimate that our overall solubility product would be underestimated. This is not what was observed, however, under-titration was unlikely given the deep-blue color of our product solutions. The other source of error is human error, however, the effects of this on our solubility product cannot be predicted. Our data, however, was quite precise, but seemingly inaccurate. It was hard to come upon a literature value for Ksp for calcium benzoate that was consistent, thus, it is possible that there may not be as much error as we believe. Conclusion: The purpose of this experiment was to understand solubility equilibria and determine Ksp for calcium benzoate. Although we obtained a large error value, this experiment was successful in teaching the concepts of solubility equilibria and metal chelation. If our error is accurate, it seems we may have over-titrated by going past the equivalence point, an error that could be fixed with more focused titration. Sources Cited: 1. Zhao, Meishan and Dragisich, Vera. General Chemistry Experiments . Hayden-McNeil Macmillan Learning, 2018. 2. “Solubility Table.” Wikipedia , Wikimedia Foundation, 13 Mar. 2019, en.wikipedia.org/wiki/Solubility_table#cite_ref-2....