DATE PERFORMED: JANUARY 20, 2011
DETERMINATION H YDROXIDE
OF THE
SOLUBILITY PRODUCT CONSTANT
OF
CALCIUM
J.M.F. UAYAN DEPARTMENT OF EPARTMENT OF CHEMICAL ENGINEERING , COLLEGE OF ENGINEERING UNIVERSITY OF NIVERSITY OF THE PHILIPPINES , DILIMAN QUEZON CITY , PHILIPPINES RECEIVED FEBRUARY 17, 2011
ABSTRACT
The objective of the experiment was to determine the solubility product constant of the reaction reaction forming forming calcium calcium hydroxide hydroxide through through the use of Titration. Titration. The prepara preparation tion of calcium hydroxide was done by mixing calcium nitrate and sodium hydroxide. The calcium hydroxide hydroxide obtained obtained was then mixed mixed with eight different different media, media, and then titrated titrated with hydrochloric acid to determine their equilibrium hydroxide concentration, which were then used to calculate the solubility product constant and the solubility of calcium hydroxide. The ionic ionic strength strength of seven seven of the eight solutions solutions was then calculated calculated and plotted plotted versus the values of the solubility of calcium hydroxide. A solubility product constant of 1.52x10 -5 was found. Compared to the literature value of 5.5x10 -6, there was a 93.72% difference.
INTRODUCTION
The activity of a hypothetical species i can be calculated using the following equation: equation:
A spar sparin ingl gly y solu solubl ble e solu soluti tion on part partia iall lly y ionizes when placed in pure water and an equilibr equilibrium ium reaction reaction takes place. place. For a hypotheti hypothetical cal ionic ionic solid, solid, A xBy, it partia partially lly ioni ionize zes s and and its its equi equili libr briu ium m reac reacti tion on is shown below:
Where ai is the activity of species I; γi is the activity coefficient; [ i ] is the concentration of species i
[1] AxBy (s) ↔ xAy+ (aq) + yBx- (aq)
For ideal cases:
Just Just like like any any reacti reaction, on, it has has a reacti reaction on quotient, Q or Q sp for this type of reaction. Reaction quotient when the reaction is in equilibr equilibrium ium is called called K sp sp or the solubility product product constant. constant. The concept concept of the K sp sp, solubility product constant is needed. K sp is the equilibrium constant of equation 1. K sp is the quanti quantific ficati ation on of the relati relation onshi ship p between the solid ionic compound and its constituent ions. K sp sp is calculated by using the following equation: [2] Ksp = a(Ay+)x a(Bx-)y a(AxBy) Where a is the activity of the substance.
[3] ai = γia[i]a
γi approaches approaches 1 (diluted solutions); ai approaches approaches [ i ] a For non-ideal cases (real case): γi can be determ determine ined d using using the DebyeDebyeHuckel equation: [4] -logγi = 0.51 Z2 µ 1+ 3.3αµ Where Z is the charge of the ion; µ is the the ioni ionic c stre streng ngth th whic which h can can be calculated using equation 9; α is the effective diameter of the ion in nanometers 1
Since the activity of a pure solid is 1 and the activity of ions ≈ concentration for diluted solutions, the K sp expression can be simplified as: [5] Ksp = [Ay+]x[Bx-]y Where [Ay+] and [Bx-] are equilibrium concentrations of Ay+ and Bx- respectively. The dissolution of Calcium hydroxide is represented by the following equation. [6] Ca(OH) 2 (s) ↔ Ca2+ + 2OHThe K hydroxide can be sp of calcium expressed as: [7] Ksp = [Ca2+][OH-]2 Where [Ca 2+] and [OH-] are equilibrium concentrations of Ca 2+ and OHrespectively.
µ = ionic strength of the solution; Ci = molar concentration of each ion present in the solution; Zi = charge of each ion present in the solution. The main objective of this experiment is to determine the solubility product constant of the dissolution of calcium hydroxide and the various factors that can influence it. Using the results of this experiment, one can point out and evaluate some of these factors and how they affect the solubility product constant. RESULTS AND DISCUSSION
There are some limitations of the K sp concept. Every salt has a K sp. But if someone would write the K sp expressions of highly soluble ionic compounds, one must consider the ions’ activities, not their concentrations. There are many factors that can affect the solubility product constant. Here are some of those factors: Temperature Common-ion effect Diverse-ion effect Simultaneous equilibria •
In the first part of the experiment, the preparation of Ca(OH) 2 (s) was done by mixing 100mL 1 M Ca(NO 3)2 and 100mL 1 M NaOH. The precipitate obtained was then placed into 8 different media. In the next part of the experiment, titration using 0.1 M HCl as titrant was done so that the concentration of the OH would be determined. Using stoichiometry as a basis, it was assumed that [Ca 2+] was half of the [OH -] obtained in the titration. K sp was then calculated using equation 7. After obtaining the K sp values, solubility was then calculated using the equation: [8] s = 3Ksp4 Where: s = solubility of the calcium ion. The solubility was then plotted versus the ionic strength of the solution; the latter calculated using the following equation: [9]
µ=12i=1nCiZi2 Where:
• • •
Since K sp is a just a K eq of a reaction between ions and a solid solute in a saturated solution, K sp is affected by changes in temperature. The common-ion effect is also a factor that affects the K sp of a certain reaction. In the experiment another source of calcium ions, Ca(NO 3)2, was added. According to the Le Châtelier’s principle, the equilibrium would shift to the left so that the extra Ca 2+ would be consumed, thus this made the calculated K sp smaller than the literature value. Another factor that can affect the solubility product constant is the presence of simultaneous equilibria. Equilibrium reaction in equation 6 is never the sole process occurring. For example, another equilibrium reaction: [10] Ca(OH)2 (aq) ↔ Ca2+ + 2OH There are many equilibrium reactions that may compete with equation 6, thus calculations based on the K sp concept may 2
fail drastically if these simultaneous equilibria aren’t taken into account. The table below shows the diverse-ion effect‘s influence on the calculated K sp. Table I.A. The Diverse-ion effect
Medium VHCl(mL) [OH-]eq [Ca2+]eq K sp 0.500 M KCl 12.8 0.0512 0.0256 -5 6.71x10 0.100 M KCl 10.0 0.0400 0.0200 3.20x10-5 0.050 M KCl 8.50 0.0340 0.0170 1.97x10-5 0.010 M KCl 5.60 0.0224 0.0112 -6 5.62x10 0.005 M KCl 8.90 0.0356 0.0178 2.26x10-5 0.001 M KCl 9.20 -5 0.0368 0.0184 2.49x10 Distilled H2O 7.80 0.0312 0.0156 1.52x10-5 From table I.A., it can be observed that as the concentration of the spectator ions increase, the calculated solubility also increases. This can be explained by the diverse-ion effect which states that the concentrations of non-participating ions or the “spectator” ions which in this case are K + and Cl–, influence the activities (effective concentrations) of the participating ions in such a way that when the total ionic concentrations became higher, ionic interactions became too significant to be negligible. When this happens, the activities or the effective concentrations of the ions participating in the equilibrium reaction become smaller compared to their calculated (stoichiometric) concentrations because K + ions tend to surround OH - ions and Cl ions tend to surround Ca 2+ ions separating them in the process thus resulting in slightly higher solubility and higher calculated solubility product constant. From table I.A., however, it can be also observed that there were some inconsistencies on the results. This is maybe due to over-titration of some analytes which could lead to wrong measurements of HCl and wrong calculations. Also the precipitate, Ca(OH) 2 (s), was not dried before mixing it with different media. The wet Ca(OH) 2 (s) may
contain excess OH - which the titrant HCl couldn’t distinguish from the OH- that came from reaction 6. Data from 0.005 M KCl and 0.001 M KCl provided some inconsistencies. Also, one can say that the preparation and of the solutions was done incorrectly because the experimentally obtained K sp values were quite far from each other (5.62x10-6 – 6.71x10-5) and also they deviated from the literature value. The table below shows the common-ion effect‘s influence on the calculated K sp. Table I.B. The Common-ion effect Medium VHCl(mL) [OH-]eq [Ca2+]eq Solubility
Distilled H2O 7.80 0.0312 2+ 0.0156 0.100 M Ca 6.10 0.1122 0.0122
0.0156 0.0244
From table I.B., it can be observed that solubility calculated when Ca(OH) 2(s) was dissolved in Ca(NO 3)2, a solvent containing Ca2+ is smaller compared to the solubility calculated when Ca(OH) 2(s) was dissolved in distilled water. The results confirmed the prediction of the common-ion effect. The common-ion effect states that if a common ion is present before the equilibrium had taken place, which in this case Ca2+, will hinder the production of more Ca2+ and OH- and shift the equilibrium towards the production of more Ca(OH) 2(s) thus resulting in a lower K sp and solubility of Ca 2+. Table II. The values of the solubility of Ca2+ vs. ionic strength in different media.
Medium Ionic strength (µ) Solubility (M) 0.500 M KCl 0.5768 0.0256 0.100 M KCl 0.1600 0.0200 0.050 M KCl 0.1010 0.0170 0.010 M KCl 0.0436 0.0112 0.005 M KCl 0.0584 0.0178 0.001 M KCl 0.0562 0.0184 Distilled H2O 0.0468 0.0156
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From table II, it was observed that as the concentration of the spectator ions increases, the solubility of Ca2+ also increases.
bromthymol blue) instead of phenolphthalein because the pH range where phenolphthalein is effective is slightly higher than 7 (around 8-10). ANSWERS TO QUESTIONS
To further show the relationship between the solubility of Ca 2+ and the ionic strength of the solution, a graph is shown in the figure below.
Figure I. The Graph of Solubility vs. Ionic Strength
From figure 1, it was observed that as the ionic strength of solution increases, the solubility of calcium ions also increases but up only to a certain degree. The graph confirmed the predictions of the diverseion effect. Data points from 0.005 M KCl and 0.001 M KCl were removed. This is done so because they provided some inconsistencies and also the trend would be more observable. CONCLUSIONS RECOMMENDATIONS
AND
There were many important observations made. First, the significance of the medium in which the Ca(OH)2 was placed is discussed. It was observed how different media affects the solubility of Ca 2+. It was observed that if the total ionic strength of the medium increases, so would the solubility of Ca2+ but only up to some degree. Titration with HCl allowed the calculation of [OH-] and in consequence, the solubility and the K sp value. However, since the calculated solubility values and K sp’s obtained in the experiment is based on many assumptions and approximations; they deviated from the literature value which is based on the activities of the participating ions rather on their stoichiometric concentrations. Also the large percent difference can be explained by overtitration of some of the analytes.
1. The solubility of calcium hydroxide increases slightly as the ionic strength of the solution increases. The effect of the total ionic strength of the solution on the solubility of Ca 2+ is called the diverse-ion effect or salt effect. The K sp should be calculated using the activities of the participating ions not their concentrations to avoid this salt effect. 2. Those K sp values obtained from the solutions with KCl are generally larger than that of the literature value while the solubility obtained from the solution with Ca(NO 3)2 is lower than that of the literature value. However the K sp value obtained from distilled H 2O saturated with Ca(OH)2 is slightly larger than the literature value. This may be due to overtitration of the analyte with HCl which leads to the miscalculation of the concentration of OH - and in consequence, the K sp. 3. The solubility calculated from the solution in the third part (part c) of the experiment is smaller compared to the literature value and to those calculated from the solutions in the second part (part b) of the experiment. This is due to the common-ion effect. The presence of a “common” ion before which in this case Ca2+, hinders the production of more Ca 2+ (Le Chatelier’s principle) and shifts the equilibrium so that the production of more calcium hydroxide solid is favored. APPLICATIONS •
• •
•
It is recommended that one should use a pH meter or another indicator (e.g.
Application of the common-ion effect in order to prevent a tablet from dissolving. Purification of Table Salt (i.e. NaCl) Production of sodium bicarbonate (baking soda) Applications in quantitative analyses (e.g. estimation of Ba 2+ in BaSO4(s)) 4
REFERENCES
The K sp expression:
Textbooks:
Ksp = [Ay+]x[Bx-]y
[1] Petrucci, P.H., Herring, F.G., Madura J.D. and Bissonnette C.: General Chemistry, Principles and Modern Application, 10th Ed. Prentice Hall, New Jersey, 2010
Where [Ay+]x and [Bx-]y are equilibrium concentrations of Ay+ and Bx- respectively.
[2] Rosenberg, J.L., Epstein, L.M.: College Chemistry, 7th Ed. McGraw-Hill, Inc, New York, 1990 [3] Brown, T.L., Le May Jr., H.E., Bursten, B.E.: Chemistry the Central Science, 11 th Ed. Prentice Hall, New Jersey, 2010. [4] Mortimer, C.E.: Chemistry, 6 th Ed. Wadsworth Publishing Company, 1986 [5] Kimsley, V.S.: Introductory Chemistry, 2nd Ed. Brooks/Cole Publishing Company, California. 1995
Solubility of Calcium ion:
s = 3Ksp4 Where: s = solubility of the calcium ion; K sp = the calculated K eq of the dissolution of Ca(OH)2; Ionic Strength of a Solution:
µ=12i=1nCiZi2 Where: µ = ionic strength; Ci = molar concentration of each ion present in the solution; Zi = charge of each ion present in the solution.
Journals: [1] Izquierdo, A., Ferre, M., Guasch, J., Rius, F., 1985. “Determination of the Solubility Product Constants of Ni (II) and Cu (II) Complexes with N-Substituted Hydrazinedithiocarboxlic Acids”. Pergamon Journals Ltd, Volume 5, Number 4, pp. 1007-1011 [2] Mauchauffée, S., Meux, E. Schneider, M, 2008. “Selected Precipitation of Cadmium from Nickel Cadmium Sulphate Solutions using Sodium Decanoate”. Separation and Purification Technology, Volume 62, Number 2, pp. 394-400 [3] Mauchauffée, S., Meux, E., 2007 “Use of Sodium decanoate for selective precipitation of metals contained in industrial wastewater”. Chemosphere, Volume 69, Number 5, pp. 763-768 APPENDIX A. WORKING EQUATIONS
The Concentration solution:
of Hydroxide in a
[OH-] = MHCl × VHCl VTotal Where: MHCl = Molarity of the HCl used; VHCl = Volume of the HCl titrated; V Total = Total volume of the analyte Percent Error:
% = | actual – theoretical |actual × 100% Where: % = percent error Percent Difference:
% = |actual – theoretical theoretical2 × 100%
|actual
+
Where: % = percent difference B. SAMPLE CALCULATIONS
5
The Concentration of Hydroxide in distilled H2O:
ZCa2+ = +2 COH- = 0.0512 M ZOH- = -1
[OH-] = MHCl × VHCl VTotal MHCl = 0.1 M VHCl = 6.1 mL V Total = 25.0 mL
µ=120.5(1)2+ 0.5-12+0.0256(2)2+(0.0512) (-1)2 µ=0.5768 M
[OH-] = 0.1M ×7.8 mL 25 mL
Percent Error:
OH-= 0.0312 M
% = | actual – theoretical |actual × 100%
The K sp for Ca(OH) 2 in distilled water:
Ksp = Ca2+OH-2 [OH-] = 0.0312 M [Ca2+] = 0.0156 M
Actual = 1.52x10-5 (The K sp obtained from dist. H2O) Theoretical = 5.5x10 -6 (The literature value)
% = | 1.52 ×10-5– 5.5×10-6 |1.52 ×10-5 × 100%
Ksp = 0.03120.01562 %= .6382 ×100% Ksp = 1.52 ×10-5 %= 63.82% Solubility of Calcium ion: Percent Difference:
s = 3Ksp4 K sp = 1.52x10-5
s = 3 1.52 ×10-54 s = 0.0156 M Ionic Strength of a Solution (0.5 KCl medium):
µ=i=1nCiZi2 CK+ = 0.5 M ZK+ = +1 CCl- = 0.5 M ZCl- = -1 CCa2+ = 0.0256 M
% = |actual – theoretical theoretical2 × 100%
|actual
+
Actual = 1.52x10-5 (The K sp obtained from dist. H2O) Theoretical = 5.5x10 -6 (The literature value)
% = |1.52 ×10-5– 5.5×10-6 |1.52 ×10-5 + 5.5×10-62 × 100% % = .9372 × 100% % = 93.72 %
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