SOLUBILITY EQUILIBRIUM OF C ALCIUM H YDROXIDE J. M. R. CIRIO D EPART MENT OF M INING , M ETALLURGICA I NING , M ETALL URGICAL L AND AN D M ATERIAL ATER IAL E NGINE ERING , C OLLEGE OLLEGE OF E NGINE ERING U NIVERSITY , D ILIMAN , Q , QUEZON C ITY , P HILIPPINES NIVE RSITY OF T HE P HILIPPINES HILIPPINES ILIMAN ITY HILIPPINES D ATE PERFORMED : M ARCH 29, 2016
ABSTRACT ABSTRACT This experiment aims to determine the solubility product constant (K spsp ) ) of calcium hydroxide by titrimetry and the effect of temperature to the molar solubility and the solubility product co nstant, as well [1] . To assess the effect of temperature on the solubility of Ca(OH)2 , three media were used: distilled water at room temperature (30 °C), at a heated t emperature (71 °C) and at a cold water bath (9°C). To be able to calculate the K sps ,p , the saturated solutions solutions were filtered and 15mL of the aliquot aliquot were - titrated using 0.1M hydrochloric acid. The obtained data were then used to calculate for the concentrations of OH and Ca 2+, wherein the value of the latter (Ca 2+ ) also corresponds to the molar solubility of the compound [2] . The molar solubility was then used to calculate the Ksp of the different media. Using the temperature and the Ksp of the media A, B and C, a linear plot (lnK sp vs. 1/T) was constructed with the equation y=105.88x-11.771 and r 2 value of 0.007. The K spsp at 298K was then calculated using the function. The K spsp computed was 1.102x10 was 1.102x10-5 which has 100.38% defiance from the literature value of 5.5x10-6 . This experiment can be improved by accuracy in the saturated solution preparation as well as in titration t itration techniques.
INTRODUCTION Sparingly soluble salts are partially dissociated to form its ions and undergo an equilibrium reaction when placed in water [1]:
(s) ⇌ + (aq) aq) − ()
(1)
From this reaction, the equilibrium constant, Keq, can be also considered as the solubility product constant, Ksp, of the ionic solid. The relationship of the ionic solid and its ions, when quantified, is the Ksp [1].
[ ][ ] ⇌
For ideal cases, approaches 1 while A° is approximately equal to 1 mol L-1. + and − . Therefore, Also, a = 1 for pure solids and liquids. That’s why concentrations are also considered to be used in computing the Equilibrium constant [1].
≈
≈
In this experiment, the equation of the Ksp was written as: (2)
where + + and − are the activities of the ions, which is derived from
+ = °
where a Ay+ is the activity of Ay+, is the activity coefficient, + is the equilibrium concentration of + , and A° is the concentration for the reference state of + .
(3)
⇌ + −
(4)
Ksp is also related to the molar solubility of the sparingly soluble salt [2]. Molar solubility, (s) is defined as the number of moles of the salt dissolved to form a liter of saturated solution. By stoichiometry, a saturated solution of has:
(s)
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= 12 ∑
+ = xs ; − = ys There are specific factors which cause changes in the Molar solubility and to the Solubility product constant, also. These are the temperature, common ions presence, diverse ions presence, and solvent polarity change [1]. Temperature affects solubility because Ksp, as with all equilibrium constants, varies with temperature, T, according to the van’t Hoff equation [4]:
Δ (7) = Δ Where Δand Δ are the enthalphy and entropy changes for the reaction, repectively, and R is the gas constant, 8.314 J/K-mol.
The effect of common-ions is an application of the Le Chatelier’s principle. It states that when a common ion is added into the solution of dissociated solids, the result would be towards the precipitation of the dissolved solid [4]. The effect of adding an ion common to one already in equilibrium in a solubility reaction is to lower the solubility of the salt. This arises from th e solution’s shifting equilibrium which actively balances the reaction system in order to attain and maintain an equilibrium state. In this case, it is the addition to the product side that needs to be removed by formation of the original solid. The diverse-ion on the other hand deals with the interionic interactions of the ions present in the solution. This is the opposite of the common-ion effect. Adding an uncommon ion tend to increase solubility. As the ionic concentration or strength of the solution then rises, these interionic contacts becomes stronger until these interactions are not negligible. Due to the increase attractions, the activities of the participating ions or their effective concentration become lesser causing a considerable increase in solubility [4]
(8)
Where is the molarity of ion , is its charge, and the summation is over all ions present in the system. The change in solvent polarity can also be observed. Two substances with similar intermolecular forces are more likely to be soluble with each other “like dissolves dissolves like” [4]. In this experiment, the solubility product constant, Ksp, of calcium hydroxide, Ca(OH) 2, was obtained using titration. Titration is the process in which a solution of one reactant, the titrant, is carefully added to a solution of another reactant, and the volume of titrant required for complete reaction is measured [3]. The equilibrium reaction of Ca(OH)2 can be expressed as,
()(s) ⇌ +(aq) aq) 2−() (9) From this expression, the K of () sp sp
can be represented by,
− ⇌ + (10) where + and − are equilibrium concentrations of + and − , respectively . [5]
This experiment aims to determine the solubility product constant (K sp sp ) of calcium hydroxide by titrimetry and the effect of temperature to the molar solubility and the the solubility product constant, as well. METHODOLOGY Six small groups were formed. Each group was assigned to prepare a specific medium in a 250-mL beaker. Each medium was labeled using letters AF. Table 1 indicates the composition and their respective volumes of the different media. Table 1. Required volume for each media.
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B C
50 mL Distilled Water (Heated, 70°C) 50 mL Distilled Water (Cold water bath, 9°C) D 50 mL 0.10 M CaCl2 (RT) E 50 mL 0.50 M KCl (RT) F 45 mL Distilled Water + 5 mL 95% ethanol (RT) After preparing the media, the Ca(OH)2 solid was added to each medium while stirring vigorously until until no more more solid dissolves. There There was no specific amount of Ca(OH) 2 solid because even an independent amount of this is added, the excess solid, when it reaches the point of saturation, will be seen suspended. The stirring of the media continued for 5 minutes and they were left to suspend for another 10 minutes. The suspensions of each medium were filtered using a 3 and 1 filter paper. The receiving flasks for the suspensions prepared in media B and C were set to have the same temperature as the suspensions. This done to prevent the change in temperature when the suspensions are transferred. The aliquot of the filtrate with the amount of 15 mL was drawn and transferred to a 125-mL Erlenmeyer flask. To the aliquot, 3 drops of 1% phenolphthalein were added. It was then titrated with 0.10M HCl until the phenolphthalein phenolphthalein color vanished vanished to colorless. colorless. The titrant volume was recorded. Another trial was performed for each medium. RESULTS AND DISCUSSIONS In this experiment, the solubility of Ca(OH)2 was investigated upon. The initial part of the experiment involved the determination of the solubility product constant and molar solubility of Ca(OH)2 through titration. Solid Ca(OH) 2 was added in six different media until the solvent could no longer dissolve the solid upon further addition, indicating the saturation of the mixture. Careful filtration must be observed as to remove all the precipitate because certain errors can
colorless when in acidic solution. The equivalence point of the titration process is specified by the drop before the solution turns colorless. Using the data recorded from the experiment and the I.C.E. table, the values of the molar solubility for each medium were calculated. Table 2. Molar Solubility
Media
V titrant titrant 2nd Trial trial 4.8 5.0 4.0 3.8 3.7 4.0 4.6 4.6 4.2 4.7 2.9 3.5 1st
A (H2O RT) B (H2O H) C (H2O C) D (CaCl2)RT E (KCl)RT F (H2O+eth)
Molar Solubility 0.01633 0.01300 0.01283 0.01533 0.01483 0.01067
The values obtained for the molar solubility were then used to calculate the Ksp for the Media A, B and C. Table 3. Solubility Product Constants of Media A, B and C.
Media and Molar Ksp ( 4s3) − Temperature Solubility (s) 0.01633 1.742 A 0.01300 0.879 B 0.01283 0.845 C Using the calculated Ksp for the different temperatures, a linear plot based on the van’ t Hoff equation was constructed.
( )
(℃) (℃) ℃) (℃)
Figure 1. lnKsp vs. 1/T
ln Ksp vs. 1/T -10.8 -11 0 -11.2 -11.4 -11.6
0.001
0.002
0.003
y = 105.88x 105.88x - 11.771 R² = 0.007
0.004
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Using the function constructed, the Ksp at 298K was calculated with the value of − And with the literature value of −[1], the percent error reached 100.38%.
1.10210 . 5.510
Δ Δ
The experimental value of the and were also calculated using the same function and using the van’t Hoff equation . The values of the and are -880.286 J/mol and -97.864 J/Kmol, respectively. The literature value of the is [6] -986.17J/mol , which means that when calculated the deviance is 10.74% while the literature value for the is 83.39 J/K-mol [6], which means that when computed, the deviance is 217.36 %. The reason why the percent error for the entropy was so large is that although the dissolution of the ions Ca2+ and 2OH- increases the entropy as you might expect, the ordering of the water molecules around these ions decreases the entropy. It turns out that in this case the latter entropy effect dominates, so that overall the entropy decreases.
Δ
Δ
Δ
Δ
The errors could have risen from: the impurities of the reagents, inconsistent ambient temperature, poor titration techniques, change in temperature (from the cold water bath or from the hotplate to the titration process, there may have been a large change in temperature) and etc. CONCLUSIONS AND RECOMMENDATIONS The conducted experiment was both a success and a failure. An experimental K sp sp yield of − in water was attained with a 100.38% error from the literature value of 5.5x10-6 [1].
1.10210
This experiment was performed using the titration technique. Using titration, the molar solubility of the Ca(OH) 2 of different media were calculated. Thus, the Ksp of the three specific media were also determined. However, by plotting the values of the natural logarithm of Ksp vs. 1/Temperature and applying linear regression, the equation of the line was found to be y = 105.88x-
valid because it is not even close to the ideal value which is 1. This results mean that the effect of the errors were great that they have altered the results. But on the bright side, the experimental had a small deviance which means that the equation of the line is near to the ideal equation of the line.
Δ
To gather more accurate results for the former part of the experiment, it is recommended to conduct a thorough and strict filtration and titration process so as to avoid excess precipitate. Moreover, thermal conditions should also be parallel with the set-up conditions. Finally, the transfer of reagents through pipetting could either add or deduct a small but significant amount of solvent hence the results are altered. REFERENCES [1] Petrucci, R., Herring G. Madura J., Bissonnette C. General Chemistry: Principles and Modern Applications 10th Ed. Ed. Pearson Publishing Publishing Inc. Canada.2011 [2] Chang, R. Chemistry 8 th Edition Student Study Guide. The McGraw-Hill Companies. 2001. [3] Whitten, K. et. al Chemistry. Brooks/Cole, Cengage Learning: Canada, 2010. [4] Solubility and Factors affecting it. Retrieved on April 4, 2016 from http://chemwiki.ucdavis.edu/Core/Ph http://chemwiki.ucdavis.edu/Core/Physical_Che ysical_Che mistry/Equilibria/Solubilty/So mistry/Equilibria/Solubilty/Solubility_and_Fac lubility_and_Facto to rs_Affecting_Solubility [5] Solubility product constant. Retrieved on April 4, 2016 from http://www.chemteam.info/Equilibrium/calcKsp-from-titration-data.html [6]Oxtoby, David W; Pat Gillis, H;
Campion, Alan (2011). Principles (2011). Principles of Modern Chemistry . p. 547
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Appendix WORKING CALCULATIONS CALCULATIONS Note: The following equations and sample calculation exhibit the calculations and equations used to solve for the data found in the data table. In the experiment, [OH ]- and [Ca2+ ] are to be solved from the titration titration data. Here, titrant (0.1 M HCl) added to the analyte solution containing the Ca(OH) 2 solution.
is the amount of
(11) (12) Table 1. Solvolysis Solvolysis of Ca(OH) Ca(OH)2 Ca(OH)2 Initial -Change -+x Equilibrium -+x
+2x +2x
Molar Solubility of Ca(OH)2 = s = x eq = K sp sp = [Ca2+ ][OH ] -]2 K eq = (x)(2x)2 = 4x3 = 4s3
(14)
SAMPLE CALCULATIONS Media
V titrant titrant 2nd Trial trial 4.8 5.0
Average
1st
A (H2O RT) B (H2O H) C (H2O C) D (CaCl2)RT E (KCl)RT F (H2O+eth) (Media A)
(13)
4.9
4.0 3.7 4.6
3.8 4.0 4.6
3.9 3.85 4.6
4.2 2.9
4.7 3.5
4.45 3.2
() )+ + = (.) .). . = 0.0326667 M = , ()
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−= = 0.013 M 2 = − s = [Ca ] = 0.013 () )+ + = (.) .). . = , () = 0.025667M −= 0.01283 M 2 = − s = [Ca ] = 0.01283M 2+
(C)
2+
() )+ + = (.) .). . = 0.0306667 M = , () − − 2 2 = = 0.015333 s = [Ca ] = 0.015333 () )+ + = (.) .). . = 0.0296667 M = , () − − 2 = = 0.0148333 s = [Ca ] = 0.0148333 () )+ + = (.) .). . = 0.0213333 M = , () − − 2 = = 0.0106667 s = [Ca ] = 0.0106667
(D)
2+
(E)
2+
(F)
2+
Ksp of (A)
3 3 -5 K sp sp = 4s = 4(0.0163333) = 1.74294x10
Ksp of (B)
3 3 -6 K sp sp = 4s = 4(0.0130000) = 8.78800x10
Ksp of (C)
3 3 -6 K sp sp = 4s = 4(0.0128333) = 8.45431x10
Media A B C
Temperature (K) 303.15 344.15 282.15
1/T (K -1) 0.003298697015 0.00290570972 0.003544214071
Ksp 1.74294x10-5 8.78800x10-6 8.45431x10-6
Y=105.88x-11.7711 Ksp at 298 K =
. −. = 1.102111052x10
-5
|ℎ−| −. ℎ−|x 100 % = .−. |ℎ| |.| x100= 100.38% Y=105.88x-11.771
%error = %error =
=
Δ = 105.8 105.88( 8(8 8.3.314 14)) = 880 880.2.286 8632 32 J/mo J/moll |ℎ−| ℎ−|x 100 % = |−.−(−.) |x100= 10.74% %error = %error = |ℎ| |−.|
105.88= -
lnKsp -10.95795975 -11.6421234 -11.68084141
