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Unformatted text preview: 10. a. Petrucci Answers, Set #10 Chapter 18 We first assume that the volume of the solution does not change appreciably when its
temperature is lowered. Then we determine the mass of CaCZO. dissolved in each solution, recognizing that the molar solubility of CaCIO‘ equals the square root of its
solubility product constant, since it is the Only solute in the solution. At 95°C FJ12x10" = 1.1 x10“ M; At 13°C: 3 = J2.7x10" = 5.2x10" M
1.1::10‘ mol Cac,o. x123.1gCan ‘ = 0.0103 CaC,0.
1 L 90111 1 mol CaC104 5.2x10"mol Pbso, x120.13 CaC,0.
1 L soln lmol Cacp. 10001113 mass of CaC104(95 '0) = 0.725 1.):
mass of CaCZO‘ (13 °C) = 0.725 Lx = 0.00483 CaCJO‘ mass of Can. precipitated = (0.010 g — 0.0048 g) x Leadﬂl) ion forms a complex ion with chloride ion. It forms no such complex ion with
nitrate ion. The formation of this complex ion decreases the concentrations of free Pb1*(aq) and free Cl' (aq) . Thus, illei1 will dissolve in the HCl(aq) up until the value of
the solubility product is exceeded. Pb1+ (aq)+3Cl' (aq):[PbClJ (aq) The solution to this problem is organized around the balanced chemical equation. Free
[NH 3] is 6.0 M at equilibrium. The size of the equilibrium constant indicates that most c0pper(ll) is present as the complex ion at equilibrium. ' . 1+ —' +
Equation. Cu (aq)+ 410113 (aq) v—— [chnaLf (aq)
Initial: 0.10M 6.00M 0M
Change(100%rxn): —0.10M. —0.40M +0.10M
Completion: 0 M 5.60 M 0.10 M
Changes: +x M +41: M —x M
Equil: x M 5.60+4xM (0.10—x) M Let’s assume (5.60 + 4x) M = 5.60 M and (0.10 ~x) M =0.10 M
21
K =[[C“(NH3)«] Ll 1x10”: 0.10—x g 0.10 a 0.10
' [0111* ][NH3 ]‘ ' x(5.60 w 4x)‘ (5.60)‘x 983$ 0.!0 =——=9.2 10"'M= C 2*
x 983.jx(l.lx10") x [ u 1 (x <4 0.10, thus the approximation was valid) 58. First we determine [Agil that can exist with this [Cl']. We know that [Cl'] will be unchanged because precipitation will not be allowed to occur. Kw =[Ag*][cr]=1.sx10"° =[Ag‘]0.100 M; [Ag‘]=1—%"11T%M=Lsx10‘° M We now consider the complex ion equilibrium. If the complex ion‘s ﬁnal concentration is
x , then the decrease in [NHJ] is 2x , because 2 mol NH 3 react to form each mol of complex ion, as follows. Ag+ (aq) + ZNH3 (sq)?_'—‘:11\g(NH3)2 1+ (aq) We can solve the Kf expresaion for x. _ x
[Ag+][NII,]1 1.8x 10‘°(l.00—~2x)2 _
.r= (1.6x 10*)(1.sx 1(}"’)(1.00—2x)2 =0.029(1.00—4.00Jr+4.00x2)=0.029—0.12.1r+0.12x2 Kf=l.6x107= 0 = 0.029 — l . 1 2x + 0.12m:2 We use the quadratic formula roots equation to solve for x. t bim_l.12i (1.12) 4x0029x0'”J‘IM’HLHO'O” x_—_——_—_.__._,__,_
20 2x0.l2 0.24 Thus, we can add 0.025 mi)! AgN03 (~ 4.4 g AgNOJ) to this solution before we see a
precipitate of AgCl(s) form. Chapter 19 6. (a) At 75‘ C, 1 mol H10 (g. 1 atm) hasagreater entropy than 1 mol H20 (1iq.,1 atm)
since a gas is much more dispersed than a liquid. 0)) 50.0 g FexH“El—E'i = 0.896 11101 Fe has a higher entropy than 0.80 mol Fe, both (5) 55.8 3 Fe at 1 atm and 5' C, because entropy is an extensive property that depends on the
amount of substance present. (c) 1 mol Br2(1iq., 1 atm, 8' C)has a higher entmpy than 1 mo] Br1(s, latm; 8‘ C) because solids are more ordered. (concentrated), substances than are liquids. and
furdiermore. the liquid is at a higher temperature. (d) 0.312 mol 502 (g, 0.110 atm, 32.5’ C) has a higher entropy than 0.284 mo] 02 (g. 15.0 arm, 22.3‘ C) for at least three reasons. First. entropy is an extensive property that depends on the amount of substance present (more moles of 802 than 02). Second,
entropy increases with temperature (temperature of SO; is greater than that for 02.
Third, entropy is greater at lower pressures (the O; has a much higher pressure).
Furthermore. entropy generally is higher per mole for more complicated molecules. II“1 (a) Negative; A liquid (moderate entropy) combines with a solid to form another solid.
(b) Positive; One mole of high entropy gas forms where no gas was present before
(1:) Positive; One mole of high entropy gas forms where no gas was present before. (d) Uncertain; The number of moles of gaseous products is the same as the number of
moles of gaseous reactants. (e) Negative: Two moles of gas (and a solid) combine to form just one mole of gas. 10. (a) We use Trouton's rule (as; = 87 J mot" K") to estimate the normal boiling point. . —l46.9 ler:ol—(—l73.5)x1 1
AB 2
Th=—.'"¥= _l _i =30§K=3.1x10 K
AS", 87 Jmol K The literature normal boiling point for pentane is 36.1 “C (= 309 K). 12. AH; = AHHBE'ZBH — AH; [135(1)] = 30.91 mm! 0.00 kamol = 30.91 lemoi AHJ... : 30.91x10’ymo1 —.— Fl _, =3.5xto=K
ASW 8711110! K . AH'
MW: W=871m01"K"or T =
T W
“P
The accepted value of the boiling point of bromine is 58.8'C = 332 K = 3.32x102 K . Thus, our estimate is in reasonable agreement with the measured value. ...
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This note was uploaded on 04/29/2011 for the course CHEM 2080 taught by Professor Davis,f during the Spring '07 term at Cornell.
 Spring '07
 DAVIS,F

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