MOOREHW7 - C'hapter as: Acids and Bases 5]. Answer: (a)...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: C'hapter as: Acids and Bases 5]. Answer: (a) 0.10 11 NH3 {13) 0.10 111x15 (c) 0.10 )I XaC'H3CDO ((1)0.10 )1 KC}— .S‘s't'otegr and Explanation: In all these comparisons. the concentrations are the same. so we can use the ionization constants front Table 16.2 to compare their strengths. The larger the K11 the more basic the solution. Ionic compounds provide cations or anions that may be bases. so watch for them. (a) NH3 is a base. (K: = 1.8 x 10-5) NaIFt’s) —u- I‘Ca'fiaq} — r-(aqj. andF‘ is a base. (Kb =1.4 x 10-“) Given equal concentrations. a solution of NH; is more basic than a solution of XaF. (b) K33(s) — 2 K-raqi — ESE—(act). and 53- is a base. (is: =1 x lo?) r<3po4tsji —} 3 [<1qu — Pof-(aq). and pofi- is a base. (ch = 2.8 >< io-J) Given equal concentrations. a solution of K38 is more basic than a solution of KEPQ. (C) M31035) 1- :qa-(aq) —No;(aq). N0; is base. (Kb: 5 x 10-15) NaCH3coorsj —:— Nat—(sq) — CH3C‘Oo-(aq). and C'I—T3C‘OO‘ is a base. (K: = 5.6 x 10-10) Given equal concentrations. a solution of C'H3C'OONa is more basic than a solution ot‘NaNOS. (d) KC'Nts) —r- K'fiaq'} — (N'Caq). and C3?" is a base. (Kb = 2.5 x 10—5) 31-13 is a base. (Kb = l.8 x 10—5} Given equal concentrations. a solution of KCN is more basic than a solution of NH;. 110. s-lttswat'.‘ (.1) H30 H30+=CT OH- (1)) H10 Xa+=crlo; H30' = on- (c) H20 H.\'o2 H30" = 5'02‘ 311" OH‘ ((1) Hit] 11’ In" ; C710— 11* OH‘ = HCID 1*- 1’ H3O" (e) H20 1’ SH; g ('1‘ 1’ H30+ =.\'H3 011- (a H20 .\'n- = OH H30" .S‘s’t'otcgr otto‘ Eitpiomttfott: Use the methods described in the solution to Question 9?. (a) HCl is a strong acid. resulting in a solution of H30' and (‘1‘. The solution contains the following molecules and ions in order ot‘decreasing concentrations: 1-130 1* 1-130" = Cl‘ '33» OH— (b) NaC'l'O4 is a soluble ionic compound resulting in a solution of Na" and C'lOf. Neither of these ions reacts with water. The solution contains the following molecules and ions in order of decreasing concentrations: H30 Na‘ = c104- H3O_ = OH- (c) I-l'NO2 is a weak acid. resulting in a solution of a large proportion of l-TNCII2 and a small proportion of H10' and NO: The solution contains the following molecules and ions in order of decreasing concentrations: H30 LocoJ 1430- = N0; 0H- (d) NaC'lO is a soluble ionic compound. resulting in a solution ot‘Na‘ and C'lO‘. The anion is a weak base and reacts with water to a small extent to make a small proportion of HClO and OH‘. The solution contains the following molecules and ions in order of decreasing concentrations: H30 Na‘ 5 czio- OH- = HC'lO H3O— (e) NH 41:21 is a soluble ionic compound. resulting in a solution of 3.1-1: ant: (‘1‘. The cation is a weak acid and reacts with water to a small extent to make a small proportion of X413 and H30". The solution contains the following molecules and ions in order of decreasing concentrations: H30 3H; 2 c1- :2» 1430— = 371-13 s»::» 0H- (fl NaOI—I is a strong base. resulting in a solution of Na' and OH‘. The solution contains the tollowing molecules and ions in order of decreasing concentrations: H10 :2» err = OH- H30— Chapter 1?: Additional. Aqueous Equilibria 32. Anni-er: (a) 5.02 (b) 4.99 (04.06 .S‘n'otegv mid Explanation: Adapt the method described in the answers to Question 24. 28. and 30. Use the abbreviation HProp for the nionoprotic propanoic acid. The .salt sodium propanoate contains the Prop— ion. (a) Find the pH using Henderson—I-L'issalbalch. The Ka ofpropanoic acid is given: 1.4 x 10—5. As described in Section 1?.1. calculate the pKa: pKa = —logKn = —1og(l .4 >< 10-5) = 4.35 The concentration of the conjugate acid‘ I-lPtop. is 0.20 M; the concentration of the conjugate base. Pnop‘. is 0.3.0 M. oso‘l |=4.as+0.18=5.02 I! pH = 4.35 +102]; ~ £0.20}. (b) Calculate the concentration of hydrogen ion. using the total volume of the solution after the addition, but before the reaction. V: 0.010 Lxm+10 ml.=11ni.T_ 1L 1.0mLx10rnoll-I3O =G-0091MH30_ 11111.1. 1L (conc. H3O") = "he strong acid neutralize; the base in the solution. C‘HSC'OO‘. Prop" (aq) — H3O'(aq) —-— l-IPi‘op — H300") "his reaction is product—favored. so we will first make as many products as possible. Using the method of limiting reactants. run the reaction towards products until one of the reactants runs out. Prop‘ (aq) H3O'(aq) HPIOplaq) initial conc. (M) 0.30 0.0091 0.20 change as reaction occurs (M) — 0.0091 — 0.0091 — 0.0091 final conc. (M) 0.29 0 0.21 The solution is still a buffer solution._ containing an acid—base conjugate pair. so determine the pH as in (a). a 1 ! a) ."0. pH=4.35—logl =4.Sf—0.14=4.99 o. | .J (c) Calculate the concentration of hydrogen ion. using the total volume of the solution after the addition. but before the reaction. \-"= 0.010 Lxm+30 m_T_=13 mL 1 L 3.0 ml. 1.0 niol H3O— (conc. 1130‘): >< ' 13 niL 1 L = 0.23 M H3O— "he strong acid neutralizes the base in the solution. Prop‘. Prop‘ (aq) — EEO—(aq) —- HProp — H200") "his reaction is product—fiu'oled. so we will first make as many; products as possible. Using the method of limiting reactants. run the reaction towards products until one of the reactants runs out. Prop‘ (aq') H3O'(aq) HPYOPan} initial conc. (M) 0.30 0.23 0.20 change as reaction occurs (M) — 0.23 — 0.23 — 0.23 final conc. (M) 0.0? 0 0.43. The solution is still a buffer solution._ containing an acid—base conjugate pair. so we can find pl—l as in (a). " on? H = 4.85 — log p “ .__o.4‘___. = 4.85 — 0.?9 = 4.06 33. Answer: (:1) Curve 2; see explanation below (b) acid 1: 1] c: T; acid 2: pH == 9.5 (c) see exploitation below [(1) see exploitation below (e) the best choice is bl'ointlwmol blue; see explanation below .S‘o'oregr and Ertpfomrfon: (a) "he effects of acid strength on the shape of the titration curve are shown in Figure 17.7’. The curve for acid ,1 _- is the curve for the weak acid. (b) The equivalence point for acid 1 is about pH T. and the equivalence point for acid 2 is about pH 9.5. (c) At the equivalence point of a strong acid and a strong base. the solution contains the cation of the strong base and the anion of the strong acid. both of which are too weak to change the pH of the water solution. T_1us the solution is neutral. and pH = T. At the equivalence point of a weak acid and a strong base. the solution contains the cation of the strong base and the anion of the weak acid. The cation is too weak to change the pH of the water solution. but the anion ot‘a weak acid is a weak base. Thus the solution is basic. and pH 3’ 3". (d) The pH of the weak acid starts at a higher value because of a snialler degree of ionization of the weak acid compared to the ionization ofa strong acid. (e) Bromthyrnol blue should be used for acid 1 and phenolphthalein should be used for acid 2. because their color changes are near the respective equivalence points. 50. Answer: C':1C103(s)+ 2 H+(aq) —-— Cai'mq) — C'D=(g) — H309) .S‘n'oregr and Eripiotmrfott: Limestone is made of calcium carbonate. The anion of this compotuid is a base and reacts with acid in rainwater to neun'alize it by a gas-forming exchange reaction (Section 5.2}. 94. .iii‘Slt-m'.‘ 2.3 x 10-4 .S‘n'oragv and Empfonanon: When exactly half of the acid in the original solution has been neutralized. that means that equal quantities of the acid and base are present in the solution. As described in the solution to Question 18. when equiniolar quantities are present. pH = pKa. pH = 3.64 = plia Ka = 10‘P3a210‘3-54 = 2.3 x 10-4 93. .lintt'ei:-'1Etpirt.=totto.tt.' Cia5(PO4)3C)l-ll:sj # 5 Cal—(ad) — 3 POE—(M1) + OH'iaq) (a) Apatite is a relatively insoluble compound. Drinking inilk containing calcium ion increases the concentration of a product in the above reaction. and. according to LeC'hatelier's principle. drives the equilibrituu toward the formation of apatite. (b) Lactic acid will react with OH‘. The subsequent decrease in OH‘ concentration will cause apatite to dissolve as equilibrium is re—established. 102. Answer: Sample A: NaHCDg; Sample B: NaOH; Sample C: NaDH and .N'aHC203: Sample D: Nazcog (other answers can also be correct) .S‘a'oregi' and Eipfdnrtrfon: We must assume that any of these substances. if present. are present in "reasonable" concentrations. Sample A: Phenolphtltalein is colorless. suggesting the pH is below 8.3. The most acidic of the choices is Nail-1C0}. and that is a likely guess for this sample's identity. (Notice: any of these compounds in solution. even NaOH. can have a pH below 8.3. if it is sufficiently dilute. that's why we have to assume reasonable concentrations.) Sample B: We interpret the evidence to say that the methyl orange changed to its acidic color as soon as it was added. That means the solution became acidic as soon as the phenolphthalein end point was reached. That suggests the titration of a strong base with a very dramatic decline to acidic pH values after the equivalence point. I11 addition. at very low pH values (below pH = 3.0L). bubbles would have been observed as the carbonate reacted to form C'O:(g'_t. All of these interpretations lead us to believe that this sample is probably NaDl—I. Sample C: Presumably. both indicators changed color rapidly. suggesting that they were true endpoints. and not just a result of pH changes. Because two different acidic end points were reached with different indicators. this sample must be a mixture of excess strong base._ NaOl—I. and one or more of the salts. either Nafi'O3 or NaHC'O3. It is not possible to distinguish which salt is present. due to the innnediate reaction of the strong base with rial-ECO; to make the same product NaJC'OB. Sample D: This sample has a two endpoints the second at exactly twice the volume of the first. suggesting that two neutralizations occurred. That means the sample may have been pure NalC'OE. which undergoes two sequential neutralization reactions during the titration. Alternative scenarios. equally plausible. would be a mixture of equiniolar quantities of NaOH and I‘Cal-IC‘O3 or a mixture of NaOH. NaECO} and Mal-1C03 with the following proportions: (cone. NaOH) = (cone. NaHC‘Osft = Hconc. NaJC‘O3ft Other proportions are also possible. Without sortie more data. or other liniits on the quantities. no more definite answers are possible. ...
View Full Document

This note was uploaded on 06/03/2008 for the course CHEM 1A taught by Professor Nitsche during the Fall '08 term at University of California, Berkeley.

Page1 / 5

MOOREHW7 - C'hapter as: Acids and Bases 5]. Answer: (a)...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online