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Exp 22 Molar Solubility numbers only

lab report
Course

Biochemistry/Lab (CHEM 3650)

163 Documents
Students shared 163 documents in this course
Academic year: 2019/2020
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LABORATORY REPORT

EXPERIMENT: Molar Solubility Lab Section: 30875 Hypothesis: Due to Le Chatelier’s Principle and Common Ion Effect, the addition of HCl titrant to the analyte of Ca(OH) 2 will shift the solution left, indicating a more acidic solution, therefore requiring less acid to titrate the solution due to common ion effect.

  1. Complete the following tables. Table 1 Determination of Concentrations

Trial Volume of Ca(OH) (mL) 2 HCl Volume of HCl titrated(mL) [OH-] (mol/L) [Ca2+] (mol/L)

1 25 mL 0 M 12 mL 2 - 1- 2 25 mL 0 M 13 mL 2-2 1- 3 25 mL 0 M 12 mL 2-2 1- Table 1: The values seen in Table 1 are the starting amount of analyte, Ca(OH)2, the molarity of titrant, HCl, and the concentration equilibrium values for [OH] and [Ca2+].

Table 2 Determination of Ksp Titration Molar Solubility ofCa(OH) 2 (M) Average Molar Solubilityof Ca(OH) 2 (M) Ksp of Ca(OH) 2 Average Ksp ofCa(OH) 2 1 1-2 M 1-2 M

2- 2 1-2 M 2-6 2- 3 1-2 M 2- Table 2: The values seen in the Table 2 are the molar solubilities found for Ca(OH) 2 , the average molar solubility, the determined trial values of Ksp and the average Ksp. STDEV: 2-7_________ %RSD: 9%_____ Ksp %Error: 58%_ (Given the known Ksp is 5-6)

Table 3 Determination of Concentrations from Common Ion

Trial Volume of Ca(OH) (mL) 2 HCl Volume of HCltitrated (mL) [OH-] (mol/L) Molar Solubilityof Ca(OH) 2 (M) 1 25 mL 0 M 4 mL 2-4 4- 2 25 mL 0 M 4 mL 2 x10-4 4- 3 25 mL 0 M 4 mL 2 x10-4 4- Table3: The values seen in Table 3 are the volume of analyte Ca(OH) 2 , the molarity of HCl used to titrant, the concentration of [OH], and the molar solubility of Ca(OH) 2.

Average Molar Solubility in presence of common ion: 4-3_______

  1. Calculations (Use equation function in Word to show your calculations. One example for each type of calculation is sufficient.) Volume of HCl Added (mL): Final buret reading volume mL minus initial buret volume mL= volume of HCL added 12 mL- 0 mL= 12 mL HCl added Moles of HCL Added (mol): mL converted to liters. Liters times molarity= mols of HCl added 25 mL= 0 L x 0 M= 6-4 mol HCl added Moles of OH- in saturated solution: it is a 1:1 ratio for OH- to HCl 6-4 mols of HCl = 6-4 mols of OH- [OH-], equilibrium (M): mols of OH- divided by volume saturated Ca(OH) 2 = [OH-] at equilibrium 0.00064/0.025=2- [Ca2+], equilibrium (M): mols of OH- at equilibrium divided by two= [Ca2+] at equilibrium 2.56x10-2/2=1- Molar Solubility of Ca(OH) 2 (M): is equal to M of [Ca2+], equilibrium (M) 1-2 Ca2+ at equilibrium = 1-2 M Molar Solubility Average Molar Solubility Ca(OH) 2 (M): Trial 1 Molar Solubility plus Trial 2 Molar solublitiy plus Trial 3 Molar Solubility divide by three= Average Molar Solubility of Ca(OH) 2 (M) 1-2 + 1-2 + 1-2 / 3= 1-

a. Will more, less, or the same amount of hydrochloric acid titrant be used for the analysis in Part A? Explain. More HCl will be needed to be used because there is more solid. Saturated solution b. Will this inadvertent transfer increase, decrease, or have no effect on the calculated solubility product for calcium hydroxide? Explain. The Ksp will be higher because the concentrations of Ca2+ and OH- will be higher. c. Will this inadvertent transfer increase, decrease, or have no effect on the calculated molar solubility of calcium hydroxide? Explain. The concentrations of Ca2+ and OH- - will be higher, therefore the molar solubility will be higher. 2. Does adding boiled, deionized water to the titrating flask to wash the wall of the Erlenmeyer flask and the buret tip increase, decrease, or have no effect on the Ksp value of the Ca(OH) 2? Explain. Adding water will have no effect to the Ksp value of Ca(OH) 2. Ksp is this case would only effected by mols or volume of Ca(OH) 2 being changed. The Ksp is independent of water.

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Exp 22 Molar Solubility numbers only

Course: Biochemistry/Lab (CHEM 3650)

163 Documents
Students shared 163 documents in this course

University: Nova Southeastern University

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LABORATORY REPORT
EXPERIMENT: Molar Solubility
Lab Section: 30875
Hypothesis: Due to Le Chatelier’s Principle and Common Ion Effect, the addition of HCl titrant to the
analyte of Ca(OH)2 will shift the solution left, indicating a more acidic solution, therefore requiring less
acid to titrate the solution due to common ion effect.
1. Complete the following tables.
Table 1 Determination of Concentrations
Trial Volume of Ca(OH)2
(mL)
[HCl]
(M)
Volume of HCl titrated
(mL) [OH-] (mol/L) [Ca2+] (mol/L)
1 25 mL 0.05 M 12.80 mL 2.56x10-2
1.28x10-2
2 25 mL 0.05 M 13.75 mL 2.75x10-2 1.38x10-2
3 25 mL 0.05 M 12.95 mL 2.59x10-2 1.30x10-2
Table 1: The values seen in Table 1 are the starting amount of analyte, Ca(OH)2, the molarity of titrant,
HCl, and the concentration equilibrium values for [OH] and [Ca2+].
Table 2 Determination of Ksp
Titration Molar Solubility of
Ca(OH)2 (M)
Average Molar Solubility
of Ca(OH)2 (M) Ksp of Ca(OH)2
Average Ksp of
Ca(OH)2
1 1.28x10-2 M
1.32x10-2 M
2.10x10-6
2.29x10-6
2 1.38x10-2 M 2.60x10-6
3 1.30x10-2 M 2.17x10-6
Table 2: The values seen in the Table 2 are the molar solubilities found for Ca(OH)2, the average molar
solubility, the determined trial values of Ksp and the average Ksp.
STDEV: ___2.22x10 -7
____________
%RSD: _____9.67%__________
Ksp %Error: _______58.36%________ (Given the known Ksp is 5.5x10-6)
1

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