Determination of the Solubility-Product Constant for a Sparingly Soluble Salt

Transcription

1 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt To become familiar with equilibria involving sparingly soluble substances by determining the value of the solubility-product constant for a sparingly soluble salt. Experiment OBJECTIVE Apparatus spectrophotometer and cuvettes 5-mL pipets (2) 75-mm test tubes (3) 150-mm test tubes (3) no. 1 corks (3) Chemicals M K2Cr M NaN mL volumetric flasks (4) buret centrifuge ring stand and buret clamp M AgN03 APPARATUS AND CHEMICALS Inorganic substances may be broadly classified into three different categories: acids, bases, and salts. According to the Bremsted-Lowry theory, acids are proton donors, and bases are proton acceptors. When an acid reacts with a base in aqueous solution, the products are a salt and water, as illustrated by the reaction of H 2S04 and Ba(OH)i: H2S04(aq) + Ba(OHh(aq) ~ BaS04(s) + 2H20(l) [1] With but a few exceptions, nearly all common salts are strong electrolytes. The solubilities of salts span a broad spectrum, ranging from slightly or sparingly soluble to very soluble. This experiment is concerned with heterogeneous equilibria of slightly soluble salts. For a true equilibrium to exist between a solid and solution, the solution must be saturated. Barium sulfate is a slightly soluble salt, and in a saturated solution this equilibrium may be represented as follows: BaS04(s) ~ Ba 2 +(aq) + 50/-(aq) The equilibrium constant expression for Equation [2] is K = [Ba 2 +][S04 2 -] c [BaS04] The terms in the numerator refer to the molar concentration of ions in solution. The term in the denominator refers to the "concentration" of solid BaS04. Because the concentration of a pure solid is a constant, [BaS04] can be [2] [3] DISCUSSION From Laboratory Experiments, Tenth Edition, John H. Nelson and Kenneth C. Kemp. Copyright 2006 by Pearson Education, Inc. Published by Prentice Hall, Inc. All rights reserved. 55

2 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt combined with Kc to give a new equilibrium constant, Ksp, which is called the solubility-product constant. Ksp = Kc[BaS04] = [Ba 2 +][so/-] At a given temperature the value of Ksp is a constant. The solubility product for a sparingly soluble salt can be easily calculated by determining the solubility of the substance in water. Suppose, for example, we determined that 2.42 x 10-4 g of BaS04 dissolves in 100 ml of water. The molar solubility of this solution (that is, the molarity of the solution) is ( 2.42 X 10-4 g BaS04)(1000 ml)( 1 mol BaS04) = 1.04 x 10-5 M 100 ml liter g BaS04 We see from Equation [2] that for each mole of BaS04 that dissolves, one mole of Ba 2 + and one mole of so/- are formed. It follows, therefore, that and solubility of BaS04 in moles/liter = [Ba 2 +] = [S04 2 -] = 1.04 x 10-5 M Ksp = [Ba 2 +][S04 2 -] = [1.04 x 10-5 ][1.04 x 10-5 ] = 1.08 x In a saturated solution the product of the molar concentrations of Ba 2 + and S cannot exceed 1.08 X If the ion product [Ba2+][S ] exceeds 1.08 X 10-10, precipitation of BaS0 4 would occur until this product is reduced to the value of Ksp Or if a solution of Na2S0 4 is added to a solution of Ba(N0 3 )z, BaS0 4 would precipitate if the ion product [Ba 2 +][SO/-J is greater than Ksp. Similarly, if we determine that the solubility of Ag2C0 3 is 3.49 x 10-3 g/100 ml, we could calculate the solubility-product constant for Ag2C0 3 as follows. The solubility equilibrium involved is and the corresponding solubility-product expression is [4] The rule for writing the solubility-product expression states that Ksp is equal to the product of the concentration of the ions involved in the equilibrium, each raised to the power of its coefficient in the equilibrium equation. The solubility of Ag2C0 3 in moles per liter is ( 3.49 x 10-3 g Ag2C03)(1000 ml)( mol Ag2C01 ) = 1.27 x 10-4 M 100 ml liter g Ag2C03 56

3 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt so that (from Equation [ 4]) and [Ag+] = 2(1.27 x 10-4 ) = 2.54 x 10-4 M Ksp = [Ag +]2[ COl-J (from Equation [ 4]) = [2.54 x 10-4 ]2[1.27 x 10-4 ] = 8.19 x To determine the solubility of Ag 2 Cr04, you will first prepare it by the reaction of AgN03 with K 2 Cr04: If a solution of AgN03 is added to a solution of K 2 Cr04, precipitation will occur if the ion product [Ag +]2[ Cr04 2 -] numerically exceeds the value of K 5 p; if not, no precipitation will occur. EXAMPLE 1 If the Ksp for Pbl 2 is 7.1 x 10-9, will precipitation of Pb1 2 occur when 10 ml of 1.0 x 10-4 M Pb(N03h is mixed with 10 ml of 1.0 x 10-3 M KI? SOLUTION: Pbl2(s) ~ Pb 2 +(aq) + 2 i-(aq) Ksp = [Pb 2 +JWJ 2 = 7.1 x 10-9 Precipitation will occur if [Pb 2 +][i-] 2 > 7.1 x 10-9 [Pb 2 +] = ( 10 ml )(1.0 X 10-4 M) 20ml = 5.0 x 10-5 M WJ = ( 10 ml )(1.0 x 10-3 M) 20ml = 5.0 x 10-4 M [Pb 2 +][r]2 = [5.0 x 10-5 ][5.0 x 10-4 ]2 = 125 x = 1.3 x Since 1.3 x < 7.1 x 10-9, no precipitation will occur. However, if 10 ml of 1.0 X 10-2 M Pb(N0 3 h is added to 10 ml of 2.0 x 10-2 M KI, then [Pb 2 +] = ( 10 ml)(l.o x 10-3 M) 20ml = 5.0 x

4 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt and WJ = (10 ml)(2.o x 10-3) 20mL = 1.0 x 10-2 M Because 5.0 x 10-7 > 7.1 x 10-9, precipitation of Pbl 2 will occur in this solution. To determine the solubility-product constant for a sparingly soluble substance, we need only determine the concentration of one of the ions, because the concentration of the other ion is related to the first ion's concentration by a simple stoichiometric relationship. Any method that we could use to accurately determine the concentration would be suitable. In this experiment, you will determine the solubility-product constant for Ag 2 Cr0 4. This substance contains the yellow chromate ion, Cr You will determine the concentration of the chromate ion spectrophotometrically at 375 nm. Although the eye can discern differences in color intensity with reasonable accuracy, an instrument known as a spectrophotometer, which eliminates the "human" error, is commonly used for this purpose. Basically, it is an instrument that measures the fraction I/Io of an incident beam of light of a particular wavelength and of intensity Io that is transmitted by a sample. (Here, I is the intensity of the light transmitted by the sample.) A schematic representation of a spectrophotometer is shown in Figure 1. The instrument has these five fundamental components: 1. A light source that produces light with a wavelength range from about 375 to 650 nm. 2. A monochromator, which selects a particular wavelength of light and sends it to the sample cell with an intensity of Io. 3. The sample cell, which contains the solution being analyzed. 4. A detector that measures the intensity, I, of the light transmitted from the sample cell. If the intensity of the incident light is Io and the solution absorbs light, the intensity of the transmitted light, I, is less than Io. 5. A meter that indicates the intensity of the transmitted light. For a given substance, the amount of light absorbed depends on the 1. concentration 2. cell or path length 3. wavelength of light 4. solvent Detector Light source Sample cell Meter (1) (2) (3) (4) (5) _. FIGURE 1 Schematic representation of a spectrophotometer. 58

5 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt Plots of the amount of light absorbed versus wavelength are called absorption spectra. There are two common ways of expressing the amount of light absorbed. One is in terms of percent transmittance, % T, which is defined as I %T = - x 100 Io As the term implies, percent transmittance corresponds to the percentage of light transmitted. When the sample in the cell is a solution, I is the intensity of light transmitted by the solution, and Io is intensity of light transmitted when the cell only contains solvent. Another method of expressing the amount of light absorbed is in terms of absorbance, A, which is defined by [5] Io A= log- [6] I The term optical density, OD, is synonymous with absorbance. If there is no absorption of light by a sample at a given wavelength, the percent transmittance is 100, and the absorbance is 0. On the other hand, if the sample absorbs all of the light, %T = 0 and A = oo. Absorbance is related to concentration by the Beer-Lambert law: A= abc where A is absorbance, bis solution path length, c is concentration in moles per liter, and a is molar absorptivity or molar extinction coefficient. There is a linear relationship between absorbance and concentration when the Beer Lambert law is obeyed, as illustrated in Figure 2. However, because deviations from this law occasionally occur, it is wise to construct a calibration curve of absorbance versus concentration. A. Preparation of a Calibration Curve WORK IN GROUPS OF FOUR TO OBTAIN YOUR CALIBRATION CURVE, BUT EVALUATE YOUR DATA INDIVIDUALLY. Using a buret, add 1, 5, 10, and 15 ml of standardized M K 2 Cr0 4 to each of four clean, dry 100-mL volumetric flasks and dilute to the 100 ml mark with 0.25 M NaN0 3. Calculate the Cr concentration in each of these solutions. Measure the absorbance of PROCEDURE A= abc Ill u i:::: ~ i: Sl..0 <:!'. 0., 0 Concentration - _.. FIGURE 2 Relationship between absorbance and concentration according to the Beer-Lambert law. 59

6 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt these solutions at 375 nm and plot the absorbance versus concentration to construct your calibration curve as shown in Figure 2. Operating Instructions for Spectronic Turn the wavelength-control knob (Figure 3) to the desired wavelength. 2. Turn on the instrument by rotating the power control clockwise, and allow the instrument to warm up about 5 min. With no sample in the holder but with the cover closed, turn the zero adjust to bring the meter needle to zero on the "percent transmittance" scale. 3. Fill the cuvette about halfway with distilled water (or solvent blank) and insert it in the sample holder, aligning the line on the cuvette with that of the sample holder. Close the cover and rotate the light-control knob until the meter reads 100% transmittance. 4. Remove the blank from the sample holder and replace it with the cuvette containing the sample whose absorbance is to be measured. Align the lines on the cuvette with the holder and close the cover. Read the percent transmittance or optical density from the meter. B. Determination of the Solubility-Product Constant Accurately prepare three separate solutions in separate 150-mm test tubes by adding 5 ml of M AgN0 3 to 5 ml of M K 2 Cr0 4. Stopper each test tube. Shake the solutions thoroughly at periodic intervals for about 15 min to establish equilibrium between the solid phase and the ions in solution. Transfer approximately 3 ml of each solution along with most of the insoluble Ag 2 Cr0 4 to 75-mm test tubes and centrifuge. Discard the supernatant liquid and retain the precipitate. To each of the test tubes add 2 ml of 0.25 M NaN0 3. Shake each test tube thoroughly and centrifuge again. Discard the supernatant liquid, then add 2 ml of 0.25 M NaN0 3 to each of the test tubes and shake the test tubes vigorously and periodically for about 15 min to establish an equilibrium between the solid and the solution. There must be some solid Ag 2 Cr0 4 remaining in these test tubes. If there is not, start over again. After shaking the test tubes for about 15 min, centrifuge the mixtures. Transfer the clear, pale yellow supernatant liquid from each of the three test tubes to a clean, dry cuvette. Measure and record the absorbance of the three solutions. Using your calibration curve, calculate the molar concentration of Cro/- in each solution. Meter Wavelength control Wavelength scale Sample holder Light control Zero adjust and power control A FIGURE 3 Spectrophotometer controls. 60

7 Determination of the Solubility-Product Constant for a Sparingly Soluble Salt Note on Calculations You are determining the Ksp of Ag2Cr04 in this experiment. The equilibrium reaction for the dissolution of Ag2Cr04 is for which Ksp = [Ag +]2[ Cro/-]. Ag2Cr04(s) ~ 2 Ag+(aq) + Cr04 2 -(aq) You should note that at equilibrium [Ag+] = 2[Cr04 2 -]; hence, having determined the concentration of chromate ions, you know the silver-ion concentration. Waste Disposal Instructions Because chromates are hazardous, all chromate solutions should be treated with care. Avoid spilling or touching these solutions. All excess K2Cr04 solution from part A should be returned to a specially marked waste container, not to the original stock solution. All chromate solutions from part B of the experiment should be placed in the same container. Likewise, all the Ag2Cr04 samples should be disposed of in the second specially marked waste container. Silver nitrate (AgN03) solution is also hazardous. Any AgN03 solution that is spilled on the skin will cause discoloration after a few minutes. All excess AgN03 solution should be returned to a third specially marked container. The other solution used in this experiment, NaN03, is harmless; any excess can be disposed of in the sinks. Before beginning this experiment in the laboratory, you should be able to answer the following questions: 1. Write the solubility equilibrium and the solubility-product constant expression for the slightly soluble salt, CaF2. 2. Calculate the number of moles of Ag+ in 5 ml of M AgN03 and the number of moles of Cro/- in 5 ml of M K1Cr If 5 ml of M AgN03 is added to 5 ml of M K1Cr04, is either Ag+ or Cr in stoichiometric excess? If so, which is in excess? 4. The Ksp for BaCr04 is 1.2 x Will BaCr04 precipitate upon mixing 10 ml of 1 x 10-4 M Ba(N03 )i with 10 ml of 1 x 10-4 M K2Cr04? 5. The Ksp for BaC03 is 5.1 x How many grams of BaC03 will dissolve in 1000 ml of water? 6. Distinguish between the equilibrium-constant expression and Ksp for the dissolution of a sparingly soluble salt. 7. List as many experimental techniques as you can that may be used to determine Ksp for a sparingly soluble salt. 8. Why must some solid remain in contact with a solution of a sparingly soluble salt in order to ensure equilibrium? 9. In general, when will a sparingly soluble salt precipitate from solution? PRE LAB QUESTIONS 61

8 NOTES AND CALCULATIONS 62

9 Name Desk Date Laboratory Instructor REPORT SHEET Determination of the Solubility-Product Constant for a Sparingly Soluble Salt EXPERIMENT A. Preparation of a Calibration Curve Initial [Cro/-1 Volume of M K2Cr04 Total volume Absorbance Molar absorption coefficient for [Cro/ Average molar absorption coefficient Standard deviation (show calculations) B. Determination of the Solubility-Product Constant Absorbance

10 Report Sheet Determination of the Solubility-Product Constant for a Sparingly Soluble Salt Average Ksp (show calculations) Standard deviation (Show calculations) QUESTIONS 1. If the standard solutions had unknowingly been made up to be M AgN0 3 and M K 2 Cr04, would this have affected your results? How? 2. If your cuvette had been dirty, how would this have affected the value of Ksp? 3. Using your determined value of K 5 p, calculate how many milligrams of Ag 2 Cr0 4 will dissolve in 10.0 ml of H The experimental procedure for this experiment has you add 5 ml of M AgN0 3 to 5 ml of M K 2 Cr0 4. Is either of these reagents in excess, and if so, which one? 5. Use your experimentally determined value of Ksp and show, by calculation, that Ag 2 Cr0 4 should precipitate when 5 ml of M AgN0 3 are added to 5 ml of M K 2 Cr

11 Report Sheet Determination of the Solubility-Product Constant for a Sparingly Soluble Salt 6. Look up the accepted value of Ksp for Ag 2 Cr0 4 in the back of your textbook. Calculate the percentage error in your experimentally determined value for Ksp 7. Although Ag 2 Cr0 4 is insoluble in water, it is soluble in dilute HN0 3. Explain, using chemical equations. 8. What is the greatest source of error in this experiment? 9. Reviewing the procedure, indicate where careless work could contribute to the error source you identified in question 8. 65

12 Report Sheet Determination of the Solubility-Product Constant for a Sparingly Soluble Salt ' c0 tj< rri N rri 0 rri 00 l"'l ' l"'l tj< C'i N C'i 0 l"'l,... OC!,... ~ ~,... ~ c:, ci ci ' c:, ci li) ci tj< ci ('<') N,... ci ci ci ajueqjosqv tj< ci..,. ~ N >< ci I CJ 0 Sd. 66

13 Answers to Selected Pre Lab Questions 1. CaF2 ~ Ca F-; Ksp = [Ca 2 +][p-]2 2. moles Ag + = ( 5 ml ml/l ) (0.004 mol/l) 1000 = 2 x 10-5 mol moles Cr04 z- = ( 10 ~ 0 :~/L }o.0024 mol/l) = 1 x 10-5 mol 3. 2AgN03 + KzCr04 ~ 2KN03 + Ag2Cr04. The stoichiometry requires 2 mol of AgN03 for each mole of K2Cr04. millimoles AgN0 3 = (5 ml)(0.004 mmol/ml) = 0.02 mmol millimoles K2Cr04 = (5 ml)( mmol/ml) = mmol and mmol K2Cr0 4 would require mmol AgN03. Hence, KzCr04 is in excess. 4. [Ba 2 +] = (10 ml/40 ml)(l x 10-4 mol/l) = 2.5 X 10-5 mol/l; [Cr ] = (10 ml/40 ml)(l X 10-4 mol/l) = 2.5 x 10-5 mol/l. Since Ksp = [Ba 2 +][Cr ] = 1.2 X and ion product = (2.5 X 10-5 )(2.5 x 10-5 ) = 6.3 x Because the ion product is more than Ksr}I precipitation would occur. 9. A sparingly soluble salt will precipitate when the ion product exceeds Ksp. 67

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