1) Ku for HX is 7.5 10-12. ' What is the ph of a 0.15 M aqueous solution of NaX? A) 1.e B) 6.0 C) 8.0 D) 12 E) 7.9 2) Calculate the poh of aa.0827 M aqueous sodium cyanide solution at 25.0 oc. Kb for CN- is 4.9 " 10-10. A) e.3 B) 8.8 C) 10 D) 1.1 E) s.2 3) The acid-dissociation constant, K4, for gallic acid is 4.57 x 10-3. \A/hat is the base-dissociation constant, K6, for the gallate ion? A) 5.43 " 10-5 B) 7.81 1,0-6 C) 2.19,102 D) 2.19 ", 70-12 E) 4.57, 10-3 4) What is the poh of a 0.0150 M solution of bariumhydroxide? A) 12.s B) 10.4 c) 1.82 D) 12.2 E) 1.52 5) \A/hich of the following acids will be the strongest? A) H2SeO4 B) H2So3 C) HSoa- D) H2SOa E) HSO3-6) A 0.0035-M aqueous solution of a particular compound has ph :2.46. The compound is _. A) a strong base B) a weak base C) a weak acid D) a strong acid E) a salt 7) l4lhich of the following ions will act as a weak base in water? A) No3- B) OHc) cl- D) CIO- E) None of the above will act as a weak base in water. 8) HA is a weak acid. \Alhich equilibrium corresponds to the equilibrium constant K6 for A-? A) i A- (aq) + H3o+ (aq) # HA (aq) + H2o (l) B)HA(aq) * H2O(l) =- U2e*(uq) * OH-(aq) C) A- (aq) + H2o (l) + HA (aq) + oh- (aq) D) HA (aq) * oh- (aq) * nzo fl) + H+ (aq) E) A- (aq) + OH- (uq) =+ HOA2- (aq) 9) l4lhich one of the following is a BrOnsted-Lowry acid? A) CH3COOH B) HNo2 C) (CH3)3NH+ D) HF E) all of the above
10) Nitric acid is a strong acid. This means that A) HNO3 dissociates completely to H+(aq) and NO3-(aq) when it dissolves in water B) HNO3 produces a gaseous product when it is neutralized C) HNO3 does not dissociate at all when it is dissolved in water -. D) aqueous solutions of HNO3 contain equal concentrations of H+(aq) and oh-(aq) E) HNO3 cannot be neutralized by a weak base HCrHro2@il? C3H5ot (aq) + H*(oq) Ko = 1.34 x t0-s I l/ Propanoic acid, HC3H5O2, ionizes in water according to the equation above. (a) Write the equilibrium-constant expression for the reaction. (b) Calculate the ph of a0.265 M solution of propanoic acid. (c) A 0.496 g sample of sodium propanoate, NaCrHrO2, is added to a 50.0 ml sample of a 0.265 M solution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the (i) The concentration of the propanoate ion, crhror- (aq), in the solution (ii) The concentration of the H*(oq) ion in the solution The methanoate ion, HCO2-@Q), reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. HCO2 @Q + HrO@ e HCO2H(oq) + OH-(aq) (d) Given that [OH-] is 4.18 x 10-6 M in a 0.309 M solution of sodium methanoate, calculate each of the (i) The value of Ku for the methanoate ion, UCO2-@A) (ii) The value of Ko for methanoic acid, HCO,H (e) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer.
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AP@ CHEMISTRY 2OO5 SCORING GUIDELINES Question 1 HC3HrO2(aq)? CrHrO2-(aq1 + U+(aq) Ko:7.34 x l0-5 Propanoic acid, HCrHrO2, ionizes in water according to the equation above. (a) Write the equilibrium-constant expression for the reaction. Notes: u [H+][C3H5O2- ] ^' - p-16;tror1 Correct expression without K, earns I point. Entering the value of Ko is acceptable. Charges must be correct to eam 1 point. One point is earned for the correct equilibrium expression. (b) Calculate the ph of a0.265 Msolution of propanoic acid. HC3H5O2(ad e CrHrO2-@a) + H+(aq) r 0.265 0-0 C-x*x+x E 0.265 - x *x tx K : lh+l[cihso2- ] - (xxx) I 9T.o:tllis earned for recognizing that [H+] and " [HCsHsOz] Q.265-x) ltclhsoz-lhavet*tu*"-lllueintheequilibrium expressron. Assumethat A.265- x = 0.265,,2 *t then 1.34 x 10-s - 0.265 (t.34x l0-sx0.265): r2 3.55x10-6:x2 r = [H*] : 1.88 x l0-3 M ph : -log[h+] - -log(1.88 x 10-3) ph = 2.725 I One point is earned for calculating [ff]. One point is earned for calculating the correct ph. Copyright O 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for AP students and parents).
AP@ CHEMISTRY 2OO5 SCORING GUIDELINES Question 1 (continued) (c) A 0.496 g sample of sodium propanoate, NaC.HrOr, is added to a 50.0 ml sample of a 0.265 Msolution of propanoic acid. Assuming that no change in the volume of the solution occurs, calculate each of the (i) The concentration ofthe propanoate ion, C3HrO2-@q) in the solution mol NaCrH so2 : 0.496 g NaCrHrO, x 1 mol NaC.HrO2 One point is earned for 96.0 gnac3h5o2 calculating the number of mol NaCrH moles of NaCrHrO, ro2: 5.17 x l0-3 mol NaCrH soz : mol C3HrOr- [c3hro2-] : mol C3H5O2-5.17x10-3mol C:HsOz- _ volume of solution 0.050 L 0.103 M One point is eamed for the molarity of the solution. (ii) The concentration of the H*(aq) ion in the solution HCtHtOr(aq) CrHsOz-@q) + H+(aq) r 0.26s o.lo3 -o I C-x+x*xl E 0.265 -x 0.103 +-x +x I r,. [H+][C3H'O2- J (xx0.103 + x) K : - One point is earned for calculating the " [HC:HsOz] Q.265-x) I _-^r.-^ value ^f rrr+l ---=---:-- of [H+1. Assume that 0.103 +r = 0.103 and0.265 - x = A.265 Ko : 7.34 x lg-s : (xxo'103) 0.265 x = [H*] = (1.34* l0-s) " o-?92 : 3.45x10-s M ' 0.103 The methanoate ion, HCO2-@q), reacts with water to form methanoic acid and hydroxide ion, as shown in the following equation. HCO2-@q) + HrO(/) e HCO2H(aq) + OH-(aq) (d) Given that [OHl is 4.18 " 10-6 Mina0.309 Msolution of sodium methanoate, calculate each of the Copyright O 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com./apstudents (for AP students and parents).
AP@ CHEMISTRY 2OO5 SCORING GUIDELINES Question 1 (continued) (i) The value of K6 for the methanoate ion, HCOr-(aq) HCO2-@q) + HrO(/) J HCO2H + OH-(aq) r0.309-0-0 C*x--fx*x E 0.309 -.x - *x *x One point is earned for substituting x : [OH-] : 4.18 " 10-6 M +.tt x 10-6forboth [OH-] and r, _ toh-lthco2hl (xxx) (4.18x10-6)2 thco2hl, and for calculating the "r - - IHcoil (030r-x) - (030r-t) talueof Ko I x is very small (4.18 x 10-6 M), therefore 0.309 -x = 0.309 *, _ (+.t8xt0 6)2 : 5.65x l0_l " 0.309 (ii) The value of Ko for methanoic acid, HCO2H K*: Kox Ko Ku, 1.00 x I 0 la I One point is eamed for calculating a value of -8lo from Y:--)/l "a Kh 5.65 x l0-ll,tte value of K6 determined in part (d)(i). Ko=1.77x10-a (e) Which acid is stronger, propanoic acid or methanoic acid? Justifo your answer. Ko for propanoic acid is 1.34 x 10-s, and Ko for methanoic acid is 1.77 x 10- a. For acids, the larger the value of K* the greater the strength; therefore methanoic acid is the stronger acid because 1.77 x 10-4 > 1.34 x 10-s. One point is earned for the correct choice and explanation based on the Kocalculated for methanoic acid in pat (d)(ii). Copyright O 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com./apstudents (for A? students and parents).