Chemistry 12 Homework Set 4

Chem 12 Non-graded port- 1Hw set 4 A. cub, b 18. B + H2O = (B)H + OH- Kb = [BH][OH-]/[B] [OH-] = sqrt(Kb * [B]) -log[OH-] = -logKb - log[B]/[B] pOH = pKb + log[B]/[B] note that since pH + pOH = 14 and pKt + pKb = 14 pH - pKt = 14 - pKb + log[B]/[B] multiply both sides by -1 ... pKa * pH = pKa - log[B]/[B] (std H-H ezu) Prepare acid TS CH3-CH2-COOH ua = 1.1x10^-5 (pu=4.84) a. 0.100M CH3CH2OH => as usual, Ka = x^2/0.1-x x = 1.14x10^-3M = {H+} pH = 2.94 x = [CH3COO-] = 8.77x10^-3M b. 0.100M CH3CH2COOH Kb = Ku/Ka = c. 7 d. pH = pKa + log(Ebase/Eacid) pH = 4.89 + log(0.1/0.1) = 4.89 -2- Hr cet 4-ng 20. #19 +0,02M HCI (102 md HCI to IL of encas 9. Ha produces 0,02 m H+ PH = 1,70 -that supresses the Tonization of the weak acid, so propanare acid will produce Sogh... let's do it anyway... • negligable Ext). ++1 0,100M ナメ ⇒x= 0,000065M... see? plt=1,70 from the HCI 5. Here, the HCl convents gorm year to y-CCLH so we have, just considering stoichiometry, 0,089-002 0,02 Mg-cost, establish equilibrium- then these casiest approach 75 to use the HH qu- + log() 4,89 + 10 = 5,49 Ad c. 1020M H+ PH = 1,70 d. the HCI convents, ORM 4-ca to q-carlt PH = 4,89 +10g() 4.71 save then as above... 9. Naot converts the weak acid to its conj. base plt part 4,89 + log() 4,29 8 a of #20, 9,020m HO POH = 1,70 As in plt = 12,30 OH=1.70 plt the presence of the weak base has a negligable effect. C. 12.30 d. PH = 4,89 + 10g(8) = 5,07 (x)(x+2010) 1.3×10-5 = 10,100-x) W200 -3- IHw set 4-ng Buffer med 19d "vesists" pt change when strong acid base is added, "HCH502"? Buaah. -CO₂H Colts = Ph so I'll write this yom I 21.5g Phcart x = 0,176 md as Phcoalt Ka = 6,4 x 10-5 pka = 4,19 122.135 37.7g Phcoz Nax Imd = 0,267md 144.115 pH = 4.19 = 4,37 19.202 1080.176 r note that Volume TS Nidevant. Pyridine "CSHEN" I'll call these pyr 24 ルール 4 = phs of pyr is 8,77, sa pku of pyritt is 5,23 = plt [pr] GAARD ৭. 0.59 6. 5,00 4.50 -973 0.19 C. 10,27 1.86 d. 5,50 5.23 =pka 9. C pka = 6.37 = 1,03 us above, plt=7,40⇒ 10 10.8 5. pka = 7,21 la = -0,058 0,88 2 rafre 1210- -4- Hur set 4-ng (confil) c. pka = 2,12 at pH=7,15, the ratio of apar 106,000 ! [hod that's not a butter, that's HYPOT. Bulters ave anly elfective at resisting around pit pka, when the PH changes cmcmbatious of acid & cong'sbase are comparable. 35. See treat answer. 56. Titratom of 100.0 ml of 9,100M HANNHz (ujdrazine!) with 0,200M HNO3 [K_{b}=3,0\times10^{-6}] 9. surtial pH U - meak base problem - [pK_{a}=8,48] HENNH₂ + HE HANNHS + HO = 9100m-x Vigy-win Jo [K_{8}=3,0\times10^{-6}=\frac{x^{8}}{(ii-x)}] [c=7.48\times10^{-4}M=(H0^{-})] [p\circ1t=3,26] [pH=10,74] X 5. + 20 mh of 0,200M HNO - probably easiest to work 0,0040 nd HNG = 4,0 mmal in moles- or milimoles! we started with Cilac mod HaNH2 x 100 yot x کالا 1000 ار = 0,0100 md H₂NNH > 10.0 mmol 4,0 mmd HNO3 convents. So we jobazine to its conj. acid end up w 4 with 4.0 mmd H₂Nity and 6.0 mnd. In a total volume of 120 ml. HENNH ... -5- Hw set 4-ng 56. b. (cont'd) This is a buffer problem - we can just use the ratio of mmol - pH = pKa + log( (mmol H₂NNH₂) / (mmol H₂NNH₃⁺) ) pH = 8.48 + log( (6.0) / (4.0) ) = 8.66 c. + 25.0 ml. 0.200M HNO₃ (total) = 5.0 mmol, so half of the hydrazine has been protonated & pH = pKa pH = 8.48 d. + 40.0 ml. = 8.0 mmol, enough to protonate 80% of the original 10.0 mmol H₂NNH₂ pH = 8.48 + log( (2.0) / (8.0) ) = 7.88 e. + 50.0 ml. = 10.0 mmol, so all the hydrazine has been protonated and this becomes a weak acid problem. we have 10.0 mmol of H₂NNH₃⁺ in a total of 150 ml. (100 ml initially + 50 from HNO₃ soln) [H₂NNH₃⁺] = (10.0 mmol) / (150 ml) = 0.067 M as usual, Ka = 3.33x10⁻⁹ = x² / (0.067-x) x = [H⁺] = 1.49x10⁻⁵ M pH = 4.83 f. +100 ml → 50 ml excess 0.050 M HNO₃ = 0.050 M H⁺ → pH = 1.30 -6- Hw set 4-ng. I'll calculate concentrations. in M; converting to mass/L. a. \[Ag_{3}PO_{4}\Rightarrow3Ag^{+}+PO_{4}^{3-}\] \[K_{sp}=1.8\times10^{-18}=[Ag^{+})^{3}[PO_{4}^{3-}]\] - for every \(Ag_{3}PO_{4}\) that dissolves, If \([PO_{4}^{3-}]=x\) , then \([Ag^{*}]=3x\) \[\prime,8\times j0^{-18}=(3\times x)^{3}(x)\] \[=27x^{4}\] \[x=i.6\times i^{-5}\mu\] \[\{PO_{4}^{5-}\}=1,6\times10^{-5}M\] get 3 silver tous, we \(Ag_{3}PO_{4}\) has gone sodu. \[(\phi_{j}*)=4,8\times10^{-5}\mathcal{M}\] b. Same approach - Caca Ca x \[K_{gp}=8.7\times10^{-9}=x^{2}\] \(x=9.3\times10^{-5}M\) CaCO_{3} has dissdred 24 \[H_{\delta z}ci_{z}}\gg H_{\delta z}^{z+}+2~c~i^{-}\] catron 2x \[K_{sp}=1.1\times10^{-18}\] \[=x\cdot(2x)^{2}\] \[=4x^{3}\] \[x=6.5\times10^{-7}M\] " 82c. [s_{r_{3}}/\rho o_{4}\rightarrow3{s_{r}}^{r+}+2~\rho o_{4}}^{\gamma-}] for even that dissolves... 3x [K_{9p}=1\times10^{-31}=[sr^{27}]^{3}{[po_{u}^{3-}]}^{2}] 85. see text ans. (=(zx)^{\prime}(zx)^{\gamma}=i\circ8\times^{5}) !!! [x=7.5\times10^{-7}M] x2 -7- HW set 4-ng 13. Phenol - HC cu, time for a shortent- here's the skeletal structure. comman - this is a type of structural representation - Cs ave understood to be at the ends of each band (unless Is shawn instead), with enough another atan It's to give them each 4 bands (or 3 bonds for -H Acidity refers to the ability of a compound compound to to sive up an - se we need to think about this process, rather than just the structure of the acid けナ localized an O CH-TH CH less ナ favorable H H mare メ Pavovable conj. bases of CISCCH and cart 7 certainly the is happier on o than c, but it C is delocalized, which stabilizes the canj, base of phenol have delocalization (vesonance) involving the" oxygens ang-thein Erca cannot be delocalized further! ↓ A- cd -8- Hw set 4-ng C. ch 6, 30. calculate the ratio of concentrations (text calls this "ex") 9. Q 9.0 x 103 - too large, so rxn must shift to the left, iie, less N2+O2 & more NO to reach equilibrium 6. \(Q=2.4\times10^{3}-\alpha f\) equilibrium c. \[Q=1_{1}1\times10^{3}-100\] much 10, must shift right. d. \[Q=212\times10^{3}\] must shift right e. \(a=4.0\times10^{\nu}\) Lett f. at equilibrium 32. \[K=\frac{[s\sigma_{i}][N\sigma]}{[s\sigma_{i}][N\sigma_{i}]}\] \(K=3.75\) more \(sc_{3}\) \[\gamma_{.}\delta\tilde{\gamma}=\frac{(0,8~c0+x)(0,8~00+x)}{(0,8~00-x)(0,8\%-x)}\] \[\sqrt{3,7}5=\frac{0.800+x}{\sigma_{1}800-x}\] \[(\circ,800)\sqrt{3,75}-x\sqrt{3,75}=0.800*x\] \[\circ,8~\circ\circ(\sqrt{3,75}-i)=x(i+\sqrt{3,8}s)\] \[x=0,255\] check: (9800+0.255) (0,800-0,255)2 = 3,75 420 than \(SO_{2}+\mu O_{2}\) at [N_{i}+0_{i}\notin2~NO] jas phase \[K=\circ,\circ\circ\circ\circ\circ=\frac{[\hat{\mu}\circ]^{z}}{[\nu_{z}][\circ_{z}]}\] initial: \(\mu_{2}:0.8m\) Ozicizature Pressunes proportional to conc. NO: O 1 -X +2x -9- Hw set 4-ng \[c_{i}e~\zeta o=\frac{(2~x)^{z}}{(o_{i}\gamma-x)(c_{i}2-x)}\] cau we get away with sajing x is small compared to a8 & 0,2 ??? Let's try it.... \[\begin{matrix}o_{i}\circ\varsigma e&=\frac{(\gamma\times\times)^{z}}{(\sigma,8)(\sigma_{i}z)}&=\frac{4\cdot\times z}{(;\epsilon)(;z)}\end{matrix}\] > \(x=0.045\) well... belter plug that back in repeat \[\sigma_{i}\circ\varsigma\circ=\frac{4~x^{z}}{(8-0.45)(.z-,\circ4S)}\] \(\Rightarrow x=0,038\) ak, one more time = 4x² 0.050 (18-038)(2-1038) → \(x=0,039\) we've dove \[P_{\mu c}=0,0>8af_{m}\] \[P_{a_{2}}=0.76afm\] \[P_{ou}=0.164fm\] 39. \[\begin{matrix}2~co_{i}&=&2~co&+\phi_{i}\end{matrix}\] \[\kappa=2.0\times10^{-6}\mu\] \[I_{u}f:\frac{z~md}{5~i}=\underline{0.40M}\] +2x +X Change: -2x at equrtis: 0.40-2x 2x -10- HW 4-ng (cartón) K= [CO2]^2[O2] / [CO2]^2 2.0×10^-6M = (2x)(x) / (0,40-2x) = 2/0,40^2 2,0×10^0M = 4x^3/0,40^2 x^2 = 8.0×10^-8 x = 0,0043 2nd iteration 2,0×10^-6M = 4x^3/(0,40-2x)^2 x^3 = 7.66×10^-8 x = 0,00042 ~ same as before -close enough [CO2] = 0,39 M [CO] = 8,4×10^-3M [O2] = 4,2x10^-3M See text aus. =