Laboratory Exercise #7 Determination of Ksp of Potassium Hydrogen Tartrate PDF

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Laboratory Exercise #7: Determination of Ksp of Potassium Hydrogen Tartrate

Introduction: In this laboratory exercise, the Ksp, solubility product constant, of KH C 4 H 4 O 6 , potassium tartrate salt, is determined. Determination of Ksp is modelled by the dissociation of a slightly soluble ionic salt in solution below: y−¿ B(aq) x+¿ +b ¿ A(aq) A a B b(s ) ⇔a ¿ b

y−¿ ¿ ¿ B Ksp = x+ ¿ ¿a ¿ ¿ A ¿ For the dissociation reaction of

KH C 4 H 4 O 6 , it is written as:

−¿ ❑(aq) ❑ ¿ K + H C4 H4 O 6 KH C4 H 4 O 6 (s) ⇔¿ +¿ K (aq) −¿ Ksp = H C 4 H 4 O 6¿ ¿¿ ¿ +¿ (aq )

The molar solubility for the tartrate salt is written as “s” if it’s in pure water. This means that the Ksp for KH C 4 H 4 O 6 is rewritten as: Ksp =

2

[ s ] [ s ]=s

The molar solubility of the hydrogen tartrate is determined using an acid-base titration since it is a monoprotic weak acid. The reaction would be: 2−¿ ❑( aq ) +H 2 O ( l) ¿ −¿ →C 4 H 4 O 6 ¿ −¿❑( aq ) +O H ¿ H C 4 H 4 O6

In order to compare the effect of the common ion, the molar solubility is determined of the tartrate salt in both water and in a 0.10 M KCl solution. For the tartrate salt in a 0.10 M KCl +¿ K (aq ) is going to be greater than the concentration of solution, the concentrate of ¿ −¿ H C 4 H 4 O 6¿ . The expression of the Ksp is written as: Ksp = [s+0.10][s] This experiment utilizes the Le Chatelier’s Principle in order to investigate the solubility constant, Ksp. It also utilizes skills in titration in order to experimentally examine the effect of the common ion within the solubility of a partially soluble ionic salt. Experimental Section: Two 125 mL Erlenmeyer flasks were cleaned and labelled as A and B. Then students proceeded to carry out the experiment for flask A. About 1 gram of potassium acid tartrate was weighed out and the mass was recorded to the 0.001 g. 50.0 mL of water was then added to the flask and stirred for 15 minutes using the magnetic stirrer. While the solution stirred, filter paper, a clean funnel, and a 100 mL beaker was set aside. After the solution was done stirring, the solution was allowed 5 minutes to settle. Then, the solution was filtered into the 100 mL beaker. 5.00 mL of the filtered solution was pipetted into a clean 50 mL Erlenmeyer flask. 2 drops of phenolphthalein indicator was then added to the solution. Then, a small Beral pipet was filled with standardized NaOH solution. A 50 mL beaker was tared and the Beral pipet was placed in the beaker to be weighed. The NaOH solution is then titrated into the tartrate salt solution by drop. When the light pink end point was reached, the NaOH Beral pipet was weighed again. This allowed the students to determine the weight of NaOH used. The procedure was repeated for an addition of two more trials in order to obtain three sets of data. The students then moved on to carry out the experiment for flask B. About 1 gram of potassium acid tartrate was weighed out and the mass was recorded to the 0.001 g. 50.0 mL of 0.100 M KCl solution was added to the flask and stirred for 15 minutes using the magnetic stirrer. The solution was allotted 5 minutes to settle. Then the solution was filtered. The filtration and titration procedures were the same as for flask A. Calculations: Solution A Calculate the number of moles of NaOH used in each titration: Trial 1: Mass of NaOH =

−3

3.72 × 10 × 1.543=5.74 × 10

−3

Moles of NaOH = (5.74 ×1 0−3)÷ 40=1.43 ×1 0−4 mol of NaOH Trial 2: 1.45× 1 0−4 mol of NaOH Trial 3: 1.47 ×1 0− 4 mol of NaOH Calculate the number of moles of Potassium Hydrogen Tartrate (KHT) salt used in each titration: Moles of KHT=Moles of NaOH Trial 1: Trial 2: Trial 3:

−4 1.43× 1 0 mol of KHT −4 1.45× 1 0 mol of KHT −4 1.47 ×1 0 mol of KHT

Determine the molar solubility of the salt in each titration: Molar Solubility=Moles of salt per litre= Moles/Volume Trial 1: −4 ((1.43 ×1 0 )÷ 5)× 1000=2.87 × 10−2 mol of KHT/L Trial 2: 2.91 × 10−2 mol of KHT/L Trial 3: 2.94 ×1 0−2 mol of KHT/L Determine the average molar solubility of the salt: −2

−2

−2

(2.87 ×1 0 )(2.91 ×1 0 )(2.94 ×1 0 ) −2 =2.90 ×1 0 mol of KHT/L 3

Determine the value of Ksp of the salt: +¿ K (aq) −¿ Ksp = H C 4 H 4 O 6¿ ¿¿ ¿ −2 [ 2.90 ×1 0 ][ 2.90 ×1 0−2 ] = 8.41× 1 0−4

Calculate the percentage error Ksp Value:

0−4 3.87 ×1 ¿ ¿ ¿❑ −4 (8.41 ×1 0 )−¿ ¿ Solution B Calculate the number of moles of NaOH used in each titration: Trial 1: 2.18 ×1 0−3 mol of NaOH Trial 2: 2.14 ×1 0−3 mol of NaOH Trial 3: 2.14 ×1 0−3 mol of NaOH Calculate the number of moles of Potassium Hydrogen Tartrate (KHT) salt used in each titration: Trial 1: 2.18 ×1 0−3 mol of KHT Trial 2: 2.14 ×1 0−3 mol of KHT Trial 3: 2.14 ×1 0−3 mol of KHT Determine the molar solubility of the salt in each titration: Trial 1: 435 ×1 0−3 mol of KHT/L Trial 2: 427 × 10−3 mol of KHT/L Trial 3: 428 ×1 0−3 mol of KHT/L Determine the average molar solubility of the salt: −3 430 ×1 0 mol of NaOH Determine the theoretical value of the solubility of KHT in 0.10 M KCl solution: 0.184 Calculate the percentage error of the molar solubility of KHT in 0.10 M KCl: 134% Results and Discussion: The purpose of this experiment was to determine the solubility constant, Ksp, and to examine the effect of the common ion within the solubility of a partially soluble ionic salt. This was done by titrating the solution with both water and 0.10 M KCl. The Ksp of the salt in water was determined to be 8.41× 1 0−4 . The Ksp of the salt in 0.10 M KCl was determined to be 430 ×1 0−3 . These values were determined using titrations to find the molar solubility. There were possible errors associated with the experiment. For example, there were method errors that could have occured. When doing the titrations, students could have over titrated the solution past the endpoint. There also could have been personal careless errors. When measuring out the solutions, the student could have put too little or too much solution. One application of the solubility product Ksp is salting out of soap. The process of salting out is known as a purification method. This process utilizes the reduced solubilities of different moles in a solution that has a high ionic strength. Since soap is sodium salt of a high fatty acid,

the process of salting out is implemented to precipitate the soup out of the solution. Conclusion: The objective of this laboratory exercise was to determine the solubility constant, Ksp, of the tartrate salt and to examine the effect of the common ion within the solubility of a partially soluble ionic salt. The Ksp of the salt in water was determined to be 8.41× 1 0−4 . The Ksp of the salt in 0.10 M KCl was determined to be 430 ×1 0−3 . References: Applications of Solubility product , ionic product , common ion effect. https://chemistryonline.guru/applications-solubility-product/ (accessed Apr 24, 2020). CHEM 108 Laboratory Manual, Department of Chemistry, Binghamton University, Binghamton, New York, 2020, pp. 155-159. Libretexts. Salting Out. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_ Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/ Real_(Non-Ideal)_Systems/Salting_Out (accessed Apr 24, 2020)....