3600_Ch9_probs - 9-3. 9-4. 9-5. 9-8. 9-9. 9-12. 9-13. [H+]...

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Unformatted text preview: 9-3. 9-4. 9-5. 9-8. 9-9. 9-12. 9-13. [H+] + 2 [Ca2+] + [Ca(HC03)+] + [Ca(OH)"'] + [K+] = [OH‘] + [Hcog] + 2[Co§‘] + [c104] [H+] = [OH‘] + [H304] + 2[sofi‘ [H+] = [OH'] + [H A o- 2‘ 3‘ (H) 2 s 4]+2[HAsO4]+3[AsO4 H—o—Als—o' O- (a) 2[MgZ‘L] + [H+] = [Br'l + [OH—l (b) 21Mg2+l + [H+l + [MgBr+] = [Br'] + [OH—l (a) 0.20M = [Mg2+] (b) 0.40M = [Br'] (C) 0.20M = [Mg2+] + [MgBr+] (d) 0.40M = [Br'] + [MgBr+] [CH3COé] + [CH3C02H] 0.1 M We ignore H+ and OH‘ because they have no effect on other species in the problem. Zn2[Fe(CN)6](s) : 22n2+ + Fe(CN)‘g 2[Zn2+] = 4[Fe(CN)‘g] [2112+] = 2[Fe(CN)‘g] KSp = 2.1 x 10'16 = [Zn2+]2[Fe(CN)‘g‘] l. Pertinent reaction: 2. Charge balance: 3. Mass balance: 4. Equilibrium constant: Now solve: If [Zn2+] = x, then [Fe(CN)%' %x (from the charge or mass balances). Putting these values into the KSp equation gives: (102630 = 2.1><10-16 => [2112+] = x = 7.5x10‘6M Charge balance: 2[Mg2+] + [H+] = [0H1 (1) Mass balance: [Mg2+] = 4.0 x 10‘8 M (2) Equilibrium: Kw = [PF] [011-] (3) For another mass balance we cannot write [OH'] = 2[Mg2+] because OH‘ comes from both Mg(OH)2 and H20 ionization. Setting [H+] = x and [Mg2+] = 4.0 x 10'8 M in Equation 1 gives [OH'] = x + 8.0 x 10'8. Putting this into Equation 3 gives Kw : (x) (x + 8.0 x 108) :> x = [H+] = 6.8 x 10'8 M. [OH‘] = x + 8.0x 10'8 = 1.5 x 10"7 M. 9-13. 9-11. 11E EL K _ = ,77 =:+ HF = —'-‘r— = EFL] [GE] l 15 Ht PH 1m [F1 + [HFTI = lfilF] = Hath] = [F] = Lawns '2 [Ca3+] [FF = [Cab] c211]: = Km = [Ca1+] = l.4x1{r3 M [F'] [HF] = 15m = 2.5:»: 133M [marge balm-cu: Invalid bra-scans: pH is. fixed- M355 Manet: my] = Email + [HPD‘h + [HzPELi]. + [Him]: {I} = {CBI'FHE =I.Txlfl"1'h{ - . {HPDE'I [UH'] K —_- — __ {22' hi [P011 [HgI’flfl [UH'] r: = . _ 3 H [Emir-1 '[ II D _ Kb: = Ebb] {4) THEE] in = [n+1 [DH] {5) Est: = IHE+13IPD§h {5} FM“ Haw-Linn: {a}, 1:3}q :4}. 1m: can write: [moi-1 = fldffi] [W] i Pt? I = _ = __ Harm] [DH] mm: H [H PD' K K PD?- [H3PD4] = i?_ 4] F _ha bi-KIEIL 4] [0H] [DH ]3 [Hm’fi] = 1!: a: mi- [Pflh [H3PU4] = 5.15: m’i [903:1 1 {Ln x 105} Using [GI-['1 = lira-m M grits [HEPCQ] = 3.515 :4 m? [rah Putting mean values inm Equalinn {I} gives My} = [961' which can he “5:111 in Equatinn [15} t1: git-'1: - 3 _EHE:L = 3 {an [Mix] + r11+1= 1E50§1+ [HSUH + [cm-1 {b} [Hm] + [mm = 21:51:11 + [14mm in} mm = mound”. [N113] 9'13. [504'] Putting in [14+] =139-25Mg1wslflsnij = 5-51:1nfi30fi‘]. Puttifig these, Willlfifi flf [Hit] and [H503 intfi [11E mass halame gim: 3.014 [r4141] + [NHfl = E{{SU§'J+5.5h-al[r3[5fll2fll [30%] = 1.0:)? LNHm Haw we use. the Hip equaflnn: Hap = LNHIJEISUE'J = I.I|.'H1]'{NI-11'J3=1 [HHQ‘] = 5.1mm INH3] = LIN-1mm] = 5.59M ...
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This note was uploaded on 12/10/2008 for the course CHEMISTRY CHEM 3600 taught by Professor Philipj.silva during the Fall '07 term at Uni. Tartu.

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3600_Ch9_probs - 9-3. 9-4. 9-5. 9-8. 9-9. 9-12. 9-13. [H+]...

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