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Relating Equilibrium, pH, and Solubility Product Constant An Introductory Chemistry Laboratory Experiment Louis J. Gotlib South Granville High School. P.O. Box 398. Creedmwr. NC 27522 Chemistry is presented often a s a series of isolated topics having no relationship. I n m y first-year chemistry classes, I use a simple procedure for determining t h e p H of a saturated solution of a hase a s a means of enabling students t o calculate a n equilibrium constant (in this case, a soluhility product constant) from their data. I n about 30 min most firstyear high school chemistry students are able to obtain results in good agreement with published values.
Equlpment and Reagents per Student Team 100-mL beaker Spatula or scoop Weighing paper Balance Stirrine rod ~istillehwater meter csoahle of resolution to 0.2 or 0.3 nH units nH -.. . ...... nr .-nH - - =oaoer (available from ish her-~MD, catalog #S-6017C (range.p~6.510.0) and #S-60171D (range 11.0-13.0), or from Sargent-Welch as Hydrion Ultrafine pH test paper, catalog #S-65262-40) Hot plate (optional) ~heknomet&(optional) Samplesto be tested: Ca(OH)z,Mg(OH)%,Ba(OH)z,and other metal (11) hydroxides such as Cu(OHh Waste disposal: The calcium and magnesium hydroxides can be rinsed down the sink with an excess of water. Other metal hydroxides are more toxic and should not be thrown in the sink, but precipitated as solids and treated as solid waste according to local regulations.
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2. Calculate the pH, [H30t] and [OH-] of the following: (a) 0.1 M NaOH (Ans. 13,l X 10-l3 M, 0.1 M) (Ans. 3.4,4.25 X lo-' (h) 0.01 M CH&OOH (K, = 1.8 X M, 2.35 X lo-" M) (e) 1 M HCL (Ans. O,1 M, 1 X lo-" M) (Ans. 11.3,5 X 10-12 M, 2 (d) 0.1 M CH3NH2(Kb= 4.4 X X M) 3. The solubility of lead chromate, PhCr04,in water is 4.4 X gl L. Calculate the solubility in moles& and calculate Ka,. (Ans. 1.36 X 10-'M, 1.85 X 1 0 - 9 4. The pH of a saturated solution of manganese hydroxide, Mn(OH)z, is 9.1. Write the equilibrium equation expression for K.,. Calculate [H30i], [OH-], and [Mn2+],and then calculate K.,. (Ans. 7.9 X 10-'OM, 1.25 X 10WM, 6.3 X 10-6M,9.8 X 1 0 - 9 5. In asaturated solution of cadmium sulfide. CdS, the [Cd7'] is 8 X LO-13 M. Calculate K.,. (Am. 6.4 X 10-?ll
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Procedure Students should wear goggles and aprons. Student teams prepare saturated solutions of the metal hydroxide to be tested by dissolving 1g of the hase in 50 mL of distilled water. After brief stirring, the mixture is allowed to stand 3 to 5 min, and the pH of the solution is tested. The pH should be measured within 5 min to minimize the effects of the dissolving of atmospheric COz. To determine the pH, place a drop or two of the solution off a stirring rod onto the pH paper. Using the value of the pH of the solution, determine the K,,. Repeat using other metal hydroxides as time permits. For a more advanced treatment, the temperature could be varied and the pH and, therefore solubility, found at several temperatures. In general, solubility increases with increasing temperature.
SamDle Data and Calculations Student data will ronsist of p H measurements. T h e pH of a saturated solution of magnesium hydroxide is 9.8. Since = -log
pH
[H30t], then
[H30+]= 10PH
(2)
By substituting t h e p H of 9.8 we obtain: [H,O+] =
= 1.6 X 10-"M
(3)
We also know t h a t [H,Ot] [OH-] = 1.0 X 10-14
(4)
and by rearranging eq 4 and substituting e q 3 into e q 4 we obtain: [OH-]
=
--
1.0x 10-l4 1.0x
[H,Oi]
1.6 X
lo-"
= 6,3
lo-''
(5)
T h e ionization expression for magnesium hydroxide is: Mg(OH),(.,
+ 20HUaq)
+ Mg2+(aq)
(6)
and
Prelaboratory Assignment Before doing this lab, students should be familiar with ionization expressions, relating pH, [H30f], and [OH-], and with solving simple solubility product constant problems.
From the ionization expression (eq 7) we can see t h a t
1. Write the equilibrium equation and from these the K,, expres-
Substituting the value of [Mg2+]from eq 8 into e q 7:
sions for the following: (a) calcium hydroxide (b) aluminum hydroxide (c) lead sulfate (d) barium phosphate
(1)
K,, = [Mg2+][OH-l2
[Mg2'l =%[OH-]
Ka, = 0.5[OH?[OH-]-2 Since [OH-] = 6.3 X value into eq 9
Volume 67
= 0.5[OH13
(7)
(8)
(9)
from e q 5 and substituting t h a t
Number 11
November 1990
937
K,, = 0.5(6.3 X
= 1.3 X
(10)
a value in good agreement (with a n order of magnitude of 10-11 t o 10-'3) with literature values ranging from 57 X to 180 X 10-13.1-3 This approach ignores the difference between concentration and activity, but that is an appropriate simplification for a first-year high school laboratory. Postlaboratory Assignment 1. What effectwould the dissolving of large amounts of atmospheric carbon dioxide have on the measured pH? On the calculated K,,? Why? Write 2. The solubilityproduct constant for Pb(0H)z is 2.8 X the K,, expression, calculate the [OH-], [H30t], and pH of a saturated solution of Pb(0Hh. Suggest a reason you did not use this compound.
938
Journal of Chemical Education
3. Suppose that instead of using distilled water, you had mistakenly used I x 10.' M KaOH. How would that have affected the solubility of the bases used? What effect w d d t hat have had on
the calculated values of K,,?
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Brescia, F. et ai. General Chemistry, 5th ed.: Harcourt Brace Jovanovich: New York. 1988: p 603 2Zumdahl,Steven S. Chemistm Heath: Lexington, MA. 1986: p 65 1 Mortimer, C. E. Chemlsby, 6th ed.; Wadswofth: Belmont, CA, 1986: n 863 --Handbook of Chemlstty and Physics, 58th ed.: Chemical Rubber Company: 1977; p 8-254.
