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edited by:
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RON DELORENZO Middle Georgia College Cochran. Georgla 31014
The Useless Tea Kettle
The Murky Pool
Robert Perklns
Robert Perkins
Unirerslly ot Brltlsk Columbla V a n c ~ v e r ,BC, Canada V6T 125
Unlr~snyof BrHIsh Colvmbla Vancouver, BC, Canada VBT 125
In areas where limestone is abundant, acidic surface water converts calcium carbonate to the more soluble calcium hvdrogen carbonate
Soluble and insoluble! Many students are quite surprised to learn that these two terms are really only qualitative descriptions of the results of solubility equilibria. I use the reaction of aqueous silver nitrate with aqueous sodium sulfide to illustrate how one can correlate the two terms with solubility product constants
-C~(HCO~)ZI~~I CaCOw + HzC031~~) When this water is boiled the calcium hydrogen carbonate breaks down forming insoluble calcium carbonate
This precipitate, known as boiler scale, collects on the heating element and, being a poor conductor of heat, reduces the ease of boiling the water. Manv tea kettles eventuallv become useless a s a rt!sult of this biild-up of calcium carbinate. Onc method of rescuing a kettle would be to dissolve awav the deposit using distilled-water. If a 2.5-L kettle contains 125 g of calcium carbonate, how many times would the kettle have to be filled to remove all of the deposit if the K., of calcium a t 25'C? carbonate is 4.8 X Initially, the solubility of calcium carbonate in pure water must be determined let
..
..
CaC03(.) = CaZ+[,) + COs2-(.q) x = [CaZ+]= [COs2-] 4.8X10@= x . x x2 = 4.8 X [Ca2+]= 6.9 X
mol .L-I
Now the amount of CaC03 present is
The solid product, silver sulfide, is an extremely insoluble salt with a K., value of 6.3 X 10-50 a t 20%. The solubilities of silver nitrate and sodium sulfide in pure water at 20°C are 2.22 X 103g-L-I and 1.88 X lo2g L-I, respectively. I ask my students to determine the mass of all three substances which could be dissolved (separately) in the college swimming pool (7.96 X 105 L) a t 20°C.
.
mass of AgN03 =
2.22 x 103g X 7.96 X 105L = 1.77 X L
lo9 g
For Ag2S,the solubility must be determined from the K , value
.. let thus,
AgzSw =+ 2 &+I,) + Sz-w K, = [Ag+]z[S2-] x = [Sz-1, 2r = [AgC] 6.3 X 10WO= ( 2 ~ ) (x) ~. 4x3 = 6.3 X r = [S2-] = 2.5 X lo-" mol L-I
.
Thus, the volume of distilled water necessary is 1.25 mol CaC08 6.9 X mol CaCOsL 1.8 X 1OP L or As the volume of the kettle is 2.5 L, the number of times the kettle must be filled is 1.8 x
lo4L
"
mass of - 2.5 X lo-'' mol AgzS X 247.9 g AgzS X7.96XlOSL AgzS L 1mol AgzS
Considering the volume of water in the pool, this is an extremely small amount of solid. T o illustrate the dramatic shift in the equilibria by a common ion, I now ask,if someone drops 5.0 mg of silver nitrate into the pool and complete mixing takes place, how many sulfide ions can be added before precipitation of silver sulfide occurs?
2.5 L or
7.2 X lo3 times X
This feahrre presents a collenion of descriptive applications and analogies designed to help students understand some of the difficult concepts frequently encountered in chemistry. Contributions b t will prcduce a greater appreciation and knowledge of political, religious. economic, historical, and scientific aspects of llle are encouraged.
1mol Ag+ 1ma1 AgN03
X
1 7.96 X 1OS L
thus, the number of 5 2 - ions will be Volume 61
Number 4
April 1964
383
"
mol 5%-
- 4.6 X L
X7.96X105LX
6.02 X
loz3 S2- ions
1 mol S2-
ions
= 22 S2- ions
A precipitate will form upon the addition of the 23rd sulfide ion, a rather dramatic shift in the equilibrium as compared to the situation in 7.96 X lo5L of pure water where the number of S2- ions would be - 2.5 X 10-17 mol S2- ions X7.96X 1 0 5 L X 6.02 X loz3 S2- ions L 1mol S2- ions = 1.2 X loL3S2- ions
hangar. Initially, the volume of the air in the hangar must he converted to liters volume = 5.5 X 107ft3X ==
1.6 X
1 ft3
lo9L
Next, the mass of vanillin required must be determined mass =
2.0 X
lo-" L
g
X 1.6 X 109L
= 3.2 X 1 0 V g
Finally, the cost of the vanillin must be determined
What A Smell! Robert Perkins
Unlverslly 01 Brltlsh Columbla BC, Canada VBT 125
Vancouver,
The Guinness Book of World Records contains a wealth of information which has lwen used previously' to pmvide inrerrsrinr! vrot~lemsfor introductorv chemistry students. The fo~lowin~buestion is typical of some of the ones which I use with my students. Vanillin (4-hydroxy-3-methoxybenzaldehyde)is listed2 as the chemical whose aroma may he detected by the human nose at the lowest concentration. The threshold limit is 2.0 X 10-l1 g per liter of air. The Goodyear Airship hangar3 in Akron, Ohio, is the largest hangar in the world, having a volume of 5.5 X 10' f@. If the current price of 500 g of vanillin is $20.45,4 determine the cost to supply enough vanillin so that the aroma could he detectable anywhere in the Goodyear Airship -
- -
' Akers, H. A., Akers, G. E., J. CHEM.EDUC., 58, 795 (1981).
'
The Guinness Bookof WorldRecords. 1979 edition. p. 176. The Guinness Book of WorldRecords. p. 255.
Aldrlch Cafalog Handbook of Fine Chemicals. 1982-83, p. 1184.
384
Journal of Chemical Education
cost = 3.2 X
gX
$20.45 500 g
- $0.0013
The calculation has even more meaning if one brings in some authentic vanillin for the students to'smell. The sensitivity of the human nose can be discussed. The instructor can show the students just how small an amount 32 mg of vanillin corresnonds to and ask them to imaeine that small amount dispersed throughout the hangar. o n t h i s basis the class will likelssav that the nose is indeed sensitke todetect the vresence-of such a small amount of vanillin. However, if onethen calculates the concentration of vanillin molecules per liter of air 0.032 g 1 mol 6.02 X loz3molecules XX concentration = 1 mol 1.6 X 109L 152.2 g = 1.9 X 10lomolecules Thus the nose requires 79 billion molecules of vanillin per liter of air before it is able to detect them! The class is now in the position of trying to decide: is the nose sensitive or not? The author would like to thank a referee for helpful suggestions on this last section.
