Petrucci Answers Set _9

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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern