Petrucci Answers Set _9

Petrucci Answers Set _9 - I.“-I I Answers to Petrucci...

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Unformatted text preview: I.“-I I? Answers to Petrucci Problems, Set #9 - Chapter 18 We determine [F‘] in saturated CaF1 , and horn that value the concentration of F' in ppm. For CaF, K” = [Caan‘r =(s)(25)’ =4s’ = 5.3x10" s= 1.1 x new The solubility in ppm is the number of grants of CaF1 in 10‘ g of solution. We assume a solution density of 1.00 g/mL. -3 OfF=loégsolnx lmL x lLsoln x1.13-(10 molCaF2 1.00gsoln IOOOmL lLsoln x 2moIF‘ x19.0gF_' =42gF_ lmolCaF, lmolF This is 42 times more concentrated than the optimum concentration of fluoride ion for fluoridation. CaF2 is, in fat, more soluble than is necessary. {ts uncontrolled use might lead to excessive F‘ in solution. We first assume that the volume of the solution does not change appreciably when its temperature is lowered. Then we determine the mass of Mg(CmI-l,,0,)2 dissolved in each solution, recognizing that the molar solubility of IMg{CwH,,O2 )2 equals the cube root of one fourth of its solubility product constant, since it is the only solute in the solution. X, =4s’ s= Bligh: AtSO'C: s=34.8x10"3;4 =i.lx10"‘ M; At25°C: s=J’3.3x10"‘14=9.4x10"M 1.1x10" 1M c H o amountof Mg(C,5H3.O,)1(50°C)=0.96SLx rm 3( “ ’1 ’) =l.1x10"mol lLsoln 9.4 x10'5mol I\IIg(le-l,,01)2 lLsoln amount of Mg(CmH3,01)z (25°C) = 0.965Lx = 0.91x10-‘mol mass of ll/Ig(CwI«lM02)1 precipitated: 535.153 Mg(C,6H3,O;)1 x 10001113 = 1.1—0.9] x104molx =llmg ( ) lmol Mg[C“I-i,,O2 )1 lg Equation: A3180, (s) : 2Ag’ (aq) + 503' (aq) Original: — 0 M 0.150 M Add solid: — ' +2xM +11: M Equil: — ZxM (0.150+x)M 2x = [Ag+ ] = 9.7 x 10'3M; x =0.0048§M K“, = [Agf [803‘] = (2x)’ (0. l 50+x) = (9.7x 10" )1 (0.150+0.0048§) = 1.5 x 10'5 24. 34. Even though BaCO3 is more soiublc than BaSO“ it will still precipitate When 0.50 M Na2CO3(aq) is added to a saturated solution of Best)" because there is a sufficient [Bah] in such a solution for the ion product [Bah ICOf'] to exceed the value of K.) for the compound. An example will demonstrate this phenomenon. Let us assume that the two solutions being mixed are of equal volume and, to make the situatiOn even more unfavorable, that the saturated Bast), solution is not in contact with solid BaSO‘, meaning that it does not maintain its saturation when it is diluted. First we determine [Bali] in saturated BaSO‘tZaq}. K”, :[BaPIsof‘]: 1.1 x10-'" =32 s= Jm 10-” = 1.0x 10"M Mixing solutions of equal volumes means that the concentration ofsolutes not common to the two solutions are halved by dilution. '5 2+ [Ba;.]=lx 1.0x10 mol BaSO‘ x lmolBa =5loxw4 M 2 1L lmolBaSO‘ , l . 2’ [€031 ]= _x0 50 11110114053203 x lmolCD3 =0'25 M 2 IL lmolNa2C03 Qm{BaC0,} = [Ba“][co,"] = (5.0x104’)(0.25) = 1.3x10" > 5.0x104 = K,{Bacoj} Thus, precipitation of BaCO1 indeed should occur under the conditions described. The solutions mutually dilute each other. We first determine the solubility of each compound in its saturated solution and then its concentration after dilution. 3 —5 K", =[Ag*]’[so."]=1.4x10"=(2s)’s= 4:3 s 4% =1_5x10-1 M 100.0 mL 100.0 mL+250.0 mL K“, = [Pb”][Cr0i']= 2.3 x 10-13 = (s){s) = s“ 250.0 mL 250.0 mL +1000 mL = 00043 M [Ag’] =0.0086 M s=~stx10'” =5.3>=10'7 M [30315015 M x [Pb’j = [0103'] = 5.3:(10'7 x = 3.3x10" M From the balanced chemical equation, we see that the two possible precipitates are PbSO. and AglCrOr (Neither PbCrO‘ nor A32304 can precipitate because they have been diluted below their saturated concentrations.) PbCrO. + Ag,so‘ "2‘ PbSO4 + AgICrO4 Thus, we compute the value of Q, for each of these compounds and compare those values with the solubility constant product value. Q” = [Pb"][so."]= [3.8x10’7)(0.0043)= 1.6x10” < 1.61.:104 = XI, for Peso4 Thus, PbSOJs) will not precipitate. Q“, = [Ag’]2[Cr0.1']= (0.0036)2 (3.3x10-’)= 2.8x10'“ > 1.1 x 10-'2 : 1g, for A3200. Thus, ASJCIO‘(S) should precipitate. We determine [Mga‘] in the solution. 0.65 g Mg (OI-I)2 X 1 mol Mg(OH)I x 1 mol Mg“ M 2* = _——— __..—=0.011M [ g l 1 L soln 58.3 g Mgwm2 1 mol Mgmu)2 Then we determine [OH‘] in the solution, and its pH. 1.8x 10"” K“,=[Mg’*][0H-]‘=1.sx10-"=(0.011)[0H']f;[on]: 00” =4.0x10"M pOH =—1og(4.0x10-‘)=4.40 pH=14.00—4.40=9.60 ...
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