122 HW_16 key

122 HW_16 key - fl, 1:, 16:1 4,...W/2 C/w Solutions of HCl...

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: fl, 1:, 16:1 4,...W/2 C/w Solutions of HCl and HZSO4 taste sour, turn litmus paper red (are acidic), neutralize solutions of bases, react with active metals to form H2(g) and conduct electricity. The two solutions have these properties in common because both solutes are strong acids. That is, they both ionize completely in H20 to form H"(aq) and an anion. (The first ionization step for HZSO4 is complete, but the second is not.) The presence of ions enables the solutions to conduct electricity; the presence of H”(aq) in excess of 1 x 10“7 M accounts for all the other listed properties. When NaOH dissolves in water, it completely dissociates to form Na“(aq) and OH‘(aq). CaO is the oxide of a metal; it dissolves in water according to the following process: Ca0(s) + H20(l) -) Ca2"(aq) + 20H‘(aq). Thus, the properties of both solutions are dominated by the presence of OH‘(aq). Both solutions taste bitter, turn litmus paper blue (are basic), neutralize solutions of acids and conduct electricity. (a) According to the Arrhenius definition, an acid when dissolved in water increases [H*]. According to the Bronsted-Lowry definition, an acid is capable of donating H*, regardless of physical state. The Arrhenius definition of an acid is confined to an aqueous solution; the Bronsted-Lowry definition applies to any physical state. (b) HCl(g) + NH3(g) -) NH4*CI“(s) HCl is the B-L (Bransted-Lowry) acid; it donates an H+ to NH3 to form NHf. NH3 is the B-L base; it accepts the H+ from HCI. 16.6 A conjugate acid has one more H“ than its conjugate base. (a) H2AsO4‘ (b) CH3NH3+ " ’ (c) HSO; (d) HBPO4 36.10} (a) racemes-(am + H200) : HacsH7os<aqi + OH-(aq) KN,” ' + (b) H206H70;(aq)+ H200) : HceH705 *(aq) + H30 (aq) (c) H306H7O5 is the conjugate acid of HZCSH705“ HC(5H7O5 ‘ is the conjugate base of HZCSH705‘ ,/ Analyze/Plan. Based on the chemical formula, decide whether the acid is strong, weak or negligible. Is it one of the known seven strong acids (Section 16.5)? Also check Figure 16.4. Remove a single H and decrease the particle charge by one to write the formula of the conjugate base. Solve: (a) weak, N02“ (d) negligible, CH3— (b) strong, HSO4' (e) weak, CHaNH2 (c) weak, PO43“ _‘ 16.16 Acid-base equilibria favor formation of the weaker acid and base. Compare the relative strengths of the substances acting as acids on opposite sides of the equation. (Bases can also be compared; the conclusion should be the same.) Base + Acid ‘75 Coniugate acid + Coniugate base (a) CHEM) + H30*(aQ) HCl(aQ) + H200) HCl is a stronger acid than H30+ (Figure 16.4) , so the equilibrium lies to the left. (b) H200) + HN02(aQ) : H30+(a<1) + N02'(aCl) H30+ is a stronger acid than HNOZ, so the equilibrium lies to the left. (0) N03’(aQ) + H200) "€— HN03(aQ) + OH‘(aq) HNO3 is a stronger acid than H20 (Solution 16.14), so the equilibrium lies to the right. i .. .x 16.22 Kw = [D*][OD']; for pure D20, [D‘] = [OD‘]; 8.9 x 10'16 = [0”]2; [D’] = [CD] = 3.0 x 104M The pH Scale .1623 Ahalyze/Plan. A change of one pH unit (in either direction) is: _ , . lH ‘] ApH — pH2 - pH1 = -(log[H ]2 — log [H 11) = -log 2 = i 1. The antilog of +1 is 10; the . . [H ’] antilog of —1 IS 1 x 10". Thus, a ApH of one unit represents an increase or decrease in [H*] by a factor of 10. Solve: (a) ApH = 12.00 is a change of 102'00 ; [H*] changes by a factor of 100. (b) ApH = 10.5 is a change of 10"50 ; [H’] changes by a factor of 3.2. 16.26 L" (a) Kw = [H*][OH‘]. If HNO3 is added to water, it ionizes to form H*(aq) and N03"(aq). This increases [H*] and necessarily decreases [OH‘]. When [H‘] increases, pH decreases. (b) On Figure 16.5, 1.4 x 10'2 M OH‘ is between pH = 12 (1 x 10'2 M OH‘) and pH 13 (1 x 10‘2 M OH‘), slightly higher than pH = 12, so we estimate pH = 12.1. By calculation: [H‘] = Kw/[OH‘] = 10 x 1044/ 1.4 X104 M= 7.1 X10’13 M pH = —log (7.1 x 10‘”) = 12.15. If pH > 7. the solution is basic. (c) pH = 6.6 is midway between pH 6 and pH 7 on Figure 16.5. At pH = 7, [H’ = 1 x 10-7; at pH = 6, [H*]=1X10‘5 =10 x1o-7. A reasonable estimate is 5 x 10”7 M H‘. By calculation: pH = 6.6,[H“]=10‘pH =10“5‘E5 >110‘7 = 3 ><10‘7 At pH = 6 [OH‘]= 1 x. 10"; at pH =7, [OH‘] =1 x10"7 =1 x104. A reasonable estimate 5 x 10‘7 M OH" . By calculation: pOH = 14.0 - 6.6 = 7.4; [OH‘] = 10—90” = 10'“ = 4 x 10"3 M OH‘. ’ ’ '\ 16.28 “n m .' For a strong base, which is completely dissociated, [OH‘] = the initial hase concentration. Then, pOH = -log [OH‘] and pH = 14 - pOH. (a) 0.0050 M KOH = 0.0050 M OH‘; pOH = -|og (0.0050) = 2.30; pH =14 - 1.30 = 11.70 m x 1mOIKOH (b) —— 0.5000 L 56.1069 KOH = 0.073254 = 0.07325 M = [OH‘] pOH = -log (0.073254) = 1.1352; pH_ =14 _ pOH = 128648 (c) Mcxvc=deVd 0.250 M Ca(OH)2 x 10.0 mL = M, Ca(OH)2 x 500 mL . L OH = mesomwmmz Mdca( )2 500.0mL Ca(OH)2(aq) —) Ca2*(aq) + 20H‘(aq) [OH'] = 2[Ca(OH)2] = 2(5.00 X 10”3 M) = 0.0100 M pOH = -log (0.0100) = 2.000; pH =14 - pOH = 12.00 mol OH ' from NaOH + mol OH ' from Ba(OH)2 (d) [OH—I‘m“ = total L sol ution (7.5 x 10‘3 M x 00300 L) + 2(0015 M x 0.0100 L) 00400 L 2.25 x 10'4 mol OH ‘ + 3.0 x104 mol OH’ = 00131 = 0.013 M OH: _ = fl [OH ]‘°‘*" 00400 L pOH = -|og (0.0131) = 1.88; pH =14 — pOH = 12.12 7 K H‘ CiH t? 16:44 “CBH702(aq> ‘1‘ H’1aq) + C5H702'(aq); K. = M [HC8H702] [H’1=[CaH702‘]=10‘2'6° = 2.09 x 104 =21 x 104 M [HCBH702] = 0.085 - 2.09 X 10'3 = 0.0829 = 0.083 M -32 m V K..=(2‘0%3%g—)=5.3x1045 @ ‘ [H+] = 10‘pH = 10‘2'7" = 1.995 X 10‘3 = 2.0 X 10'3 M _ [H1 [F '1 _ (1.995 x 10'3)2 6.8 x 104(x -1.995 x 104) = (1.995 x 104?; 6.8 x 10‘4x = 1.357 x 10* + 3.981 x 104 = 5.338 x 10*3 x = 7.85 x 104 = 7.9 x 10“3 MHF mol = M x L = 7.85 x 104 Mx 0.500 L = 3.925 x 10“3 = 3.9 x 104 mol HF Ka=6.8 x 104 y‘16.50“* HClO(aq) #— H*(aq)+ClO‘(aq) initial 0.0075 M 0 0 equil. (0.0075 — x) M x M x M Ka: [H‘][ClO‘] = x2 z x2 =30x104 [HClO] (0.0075 - x) 0.0075 ' x2 = 0.0075 (3.0 x 104); x = 1.5 x 10“5 M: [H‘] = [H301 = [ClO‘] [HClO] = 7.5 x 104 — 1.5 x 10-5 = 7.485 x 104 = 7.5 x 104 M 4.7x10‘5MH* . . . Check. a x 100 = 0,20"/ - ' ' ' 0.0075 MHCIO o ionization, the assumption is valid —\ ;, 16.68 (a) pOH = 14.00 - 9.95 = 4.05; [OH‘]=10“"°5 = 8.91 x 10‘5 = 8.9 x 10'5 M ’ ciaHZiNoaiaqi + H200) ‘1‘ cisHmNoawiaq) + OH‘(aq) initial 0.0050 M 0 0 equil. (0.0050 - 8.9 x 104) 8.9 x 104 M 8.9 x 104 M C H NO H‘ OH’ ~5 2 = ‘8 21 3 H 1: “391x10 ) =1.623<10'6=1.6)<10*"’ [C18H21N03] (0.0050 - 8.91 x 105) (b) pr = -log (Kb) = -log (1.62 x 1045) = 5.79 /. 1;: L “18,8O'HThe solution is basic because of the hydrolysis of P0431 The molarity of P043" is 50.0 g Nei3PO4 1 mol Na3PO‘1 ——____ x ————.__ = 0.3051 = 0.305 M PO43“ 1.00 L soln 183.9 g Na3PO4 P043'(aq)+ H200) x‘i HPOflaq)’ + OH‘laq) HPOZ‘ OH’ K —14 Ki,=[ 4 H ]= w =——Lx10 =0.0238=2.4 x10-2 [PO43] i<a for HPof‘ 4.2 x 10‘13 ignoring the further hydrolysis of HPO4 ‘. [OH'] = [HPOi‘l = x. [P0431 = 0.305 - x 2 2.4 x 10—2 = TUB—’85“); x2+2.4 x 10-2x-0.0073 = 0 . -X Since Kb is relatively large, we will not assume x is small compared to 0.305. X = -0.024 i1/(0.024)2 -4(1)(—0.0073) = —0.024:‘/0.0299 2(1) 2(1) x = 0.074 MOH‘; pH =14 + log (0.074) = 12.87 16.84 (a) For binary hydrides, acrd strength increases gorng across a row, so HCl is a stronger acid than HZS. (b) For oxyacids, the more electronegative the central atom, the stronger the acid, so H3PO4 is a stronger acid than H3AsO4. (c) HBrO3 has one more nonprotonated oxygen and a higher oxidation number on Br, so it is a stronger acid than HBrOz. (d) The first dissociation of a polyprotic acid is always stronger because H+ is more tightly held by an anion, so H20204 is a stronger acid than HCZO4‘. (e) The conjugate base of benzoic acid, C7H502", is stabilized by resonance, while the conjugate base of phenol, CGHSO‘, is not. HC7H5O2 has greater tendency to form its conjugate base and is the stronger acid. ;_____.—.——————— (36.92 Lewis Acid Lewis Base (a) HNo2 (or H“) OH- (b) FeBra (Fe3*) Br- (c) Zn2+ NH3 (d) so2 H20 16.93 (a) Cu”, higher cation charge (b) Fe“, higher cation charge (c) AP", smaller cation radius, same charge 16.94 (a) , ZnBrz, smaller cation radius, same charge (b) Cu(N03)2, higher cation charge (c) NiBrz, smaller cation radius, same charge ...
View Full Document

Page1 / 5

122 HW_16 key - fl, 1:, 16:1 4,...W/2 C/w Solutions of HCl...

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