Solibility Lab Report Sheeta Ella Jul 7,207
Experiment 3.A. Determining a Solubility Product Constant for PbI 2 Abstract (I don t think u need headers? But up to u) The solubility of a substance can be changed by the temperature of the system. This experiment is designed to investigate how (an increase in temperature effects the solubility of solutions.) the solubility of most solutions gets affected by an increase in temperature, and also provide an opportunity to calculate the solubility product constant (K sp ) through experimental means. The( procedure starts from the creation of the precipitate lead (II) iodide with different initial concentrations; and(by) heating it up until the precipitate dissolves at a certain temperature discovered by this experiment. The results obtained in the lab were expected, as an increase in the solubility of lead (II) iodide was observed, as well as the increase in temperature. However, the range of K sp values from the lab differ(varies) from the literature or websites that were used as a reference. Therefore, further experimentation must be performed to determine the cause of the different range in K sp values and discover a more accurate result(s),(.moreover;) as well as the procedure must be revisited and reconsidered to eliminate any mistakes made in the lab by experimental and human errors. Introduction Any solid salts are either soluble or have a low solubility. Solubility is the maximum amount of solute that can be dissolved in a given volume of solvent before it becomes saturated. Temperature usually increase the amount of solute that can be dissolved in a give volume, however, this doesn't apply to all the compounds. The solubility product constant (K sp ) is a special kind of equilibrium constant that describes the equilibrium that exists in a saturated solution which could also be changed by the temperature. Precipitate forms when the trial K sp from the experiment (Q) is equal to or greater than the K sp from the data book. Pb(NO 3 ) (aq) + 2KI (aq) PbI 2 (s) + 2KNO 3 As the equation above, Solid lead (II) iodide is created as a precipitate by mixing potassium iodide and lead (II) nitrate. Based on the data book, lead (II) iodide has a low solubility, therefore we expected that the change in solubility caused by the temperature would be larger and easier to observe and (similarly)calculate the solubility product constant. The solubility is increased when the amount of solid that gets dissolved is greater. This happens when the equilibrium shifts to the side with dissolved ions, which could be caused by the change in temperature. Thus, we hypothesized that the increased temperature dissolves solids faster and is necessary for a precipitate with higher concentrations to dissolve. Equipment and Materials
Refer to the Materials and Apparatus section of the lab handout Experiment 3.A. Determining a Solubility Product Constant for PbI 2. There were no 00mL beakers or a dropping pipette used in the lab. Experimental Procedure Refer to the Procedure section of the lab handout Experiment 3.A. Determining a Solubility Product Constant for PbI 2. Step one was skipped in the lab due to the availability of the prepared materials. Observation and Results Table : Temperature of the room: 23 C(+/-0.) Test tube A B C D E F Vol. 0.00M Pb(NO 3 ) added (ml) (+/-0.0) 0.00 8.00 6.00 4.00 3.00 2.00 Vol. Water added (ml) (+/-0.0) 0.00 2.00 4.00 6.00 7.00 8.00 Vol. 0.020M KI added (ml) (+/-0.0) 0.00 8.00 6.00 4.00 3.00 2.00 Vol. Water added (ml) (+/-0.0) 0.00 2.00 4.00 6.00 7.00 8.00 Precipitate when mixed at room temperature? YES YES YES NO NO NO Temperature at which precipitate dissolves ( C ) (+/-0.) 99. 8.5 60.8 0.006 Solubility of Lead (II) Iodine 0.005 99., 0.005 Solubility (M) 0.004 0.003 0.002 60.8, 0.003 8.5, 0.004 0.00 0 0 20 40 60 80 00 20 Tmperature ( C ) Figure. The solubility of lead (II) iodine at given temperatures with a linear line.
The concentration of [Pb 2+ ] in the final solution (Should be 0.00M by ratio) 0.00M 0.00M 0.020M by mole ratio Pb(NO 3 ) 2 Pb 2+ (aq) + 2NO 3 - (a Test tube A: [Pb 2+ ] =( "."$"% )(0.000L)=0.0050M Test tube B: [Pb 2+ ] =( "."$"% Test tube C: [Pb 2+ ] =( "."$"% Test tube D: [Pb 2+ ] =( "."$"% Test tube E: [Pb 2+ ] =( "."$"% Test tube F: [Pb 2+ ] =( "."$"% KI K + (aq) + I - (aq) ) (0.0080L) =0.0040M ) (0.0060L) =0.0030M ) (0.0040L) =0.0020M ) (0.0030L) =0.005M ) (0.0020L) =0.000M Test tube A: [I - ]= ( "."&"% )(0.000L) =0.00M Test tube B: [I - ]= ( "."&"% )(0.0080L) =0.0080M Test tube C: [I - ]= ( "."&"% Test tube D: [I - ]= ( "."&"% Test tube E: [I - ]= ( "."&"% Test tube F: [I - ]= ( "."&"% )(0.0060L) =0.0060M )(0.0040L) =0.0040M )(0.0030L) =0.0030M )(0.0020L) =0.0020M Q value for each test tube Q=[Pb 2+ ] [I - ] 2 Test tube A: 0.0050*0.00 2 = 5.0*0-7 Test tube B: 0.0040*0.0080 2 = 2.6*0-7 Test tube C: 0.0030*0.0060 2 =.*0-7 Test tube D: 0.0020*0.0040 2 = 3.2*0-8 Test tube E: 0.005*0.0030 2 =.4*0-8 Test tube F: 0.000*0.0020 2 = 4.0*0-9 Solubility of PbI 2 for test tubes that formed a precipitate. (A to C) PbI 2 Pb 2+ (aq) + 2I - (aq) K sp = [Pb 2+ ] [I - ] K sp = 4x 3 x = Ksp/4 Test tube A: 5.0 0 56 /4 = 0.0050M = [PbI 2 ] Test tube B: (2.6 0 56 )/4 = 0.0040M = [PbI 2 ] Test tube C: (, 0 56 )/4 = 0.0030M = [PbI 2 ] Discussion As our hypothesis states,our hypothesis states that an increase in temperature is often required to dissolve precipitates with higher concentrations,(.) and thus, correspondingly we expected that as the concentration of precipitate gets greater, the temperature must increase as well in order for the precipitate to dissolve. This expectation is based on the assumption that the increased temperature will increase the solubility. The results obtained in each test tube A, B and C with its different concentration of precipitate and temperature matched our expected observations, as the concentrations in test tubes A, B and C were decreased. Correspondingly, the temperature that was needed to dissolve the precipitates decreased as well. The results meansshow that increasing the temperature shifts the equilibrium to the right in this reaction; therefore, it is an endothermic in the dissociation direction. The solubility is increased due to the favoured products, one of which was the precipitate of lead (II) iodide. Our hypothesis matched the general solubility equilibria. To provide a further support, sodium arsenate,
Na 2 HAsO 4 (Boundless) has a similar trend in temperature and solubility relationship as the solubility increases as the temperature increases as well. Also, with as the theories proposed by chemical reaction kinetics (Boundless), whichit states that increasing the temperature will increase the overall rate of solubility. Inthorughout this experiment, only some solutions formed a precipitate. The experimental K sp values (Reaction Quotient, Q) of those solutions were greater than the K sp values found from the data book;which caused the precipitate to form. therefore, a precipitate was formed. The rest of solutions did not form a precipitate due to lower ion concentration than concentration of those saturated solutions with precipitate. To determine the solubility of lead (II) iodide at 23 C, we are only able to calculate the solubility of the ones that did form a precipitate. However, there was an issue with the calculated K sp values range in the data, which was.*0-7 to 3.2*0-8 in test tube C, and D. These values were different from the K sp values from the literature. This is an incorrect result because test tube C formed a precipitate but test tube D did not. Although the solutions with lesser concentration had closer K sp values to the K sp, 8.5*0-8 at 25 C from the data book or the literature s K sp, 4.4*0-9 at 20 C (Wikipedia), this trial K sp range (Q) is slightly greater than values from the literature. Which could have been cause by mistake throughout the lab.therefore, there must have been some errors or mistakes made in the lab. Those errors that caused this discrepancy could have been observational, experimental or human errors. For an experiments done with solutions like this one, it is important to avoid the contamination from unwashed equipment with unknown chemicals and the dilution of concentration from the water used to rinse the equipment. In addition, transferring the solutions from the graduated cylinder to test tubes might have changed the volume of solution slightly by leaving a small amount in the cylinder, which can cause a difference in concentration and the amount of precipitate. The lab should have been done in a closed system so unknown substances or particles in the air cannot enter the system. To make this experiment more accurate, the equipment should be washed and dried completely, preferably by paper towels so the concentration of solutions will not be diluted by the water that was used to rinse them out. Conclusion The results from this experiment supported our hypothesis that an increase in temperature will increase the solubility of a solution. This result agrees with the trend seen in sodium arsenate, Na 2 HAsO 4 (Boundless) as the relationship between temperature and solubility is similar to our experiment. However, further experimentation is required for a closer result due to the range difference in K sp values from those in reference. To improve this lab, it is important to do the procedure slowly in a closed system.
References Boundless. "Solid Solubility and Temperature." Boundless Chemistry Boundless, July 27 207. 207 from https://www.boundless.com/chemistry/textbooks/boundless-chemistrytextbook/solutions-2/factors-affecting-solubility-94/solid-solubility-and-temperature-403-538/ Wikipedia The Free Encyclopedia. Lead (II) iodide. July 25 207 from https://en.wikipedia.org/wiki/lead(ii)_iodide