AP Chemistry aq l aq aq CHNH +HO CHNH +OH CHNH

Transcription

1 1. (a) aq l aq aq HO OH OH C 6H 5NH 3 AP Chemistry 2003 () poh log[oh ] poh ph log[oh ] According to the alanced equation aove, aniline [OH ]6.6x10 6 M dissociates into equal concentrations of [OH ]6.6x10 6 M [C 6 H 5 NH 3 ] C 6 H 5 NH 3 and OH so [C 6 H 5 NH 3 ] [OH ]. 6 6 OH C 6H 5NH x 10 M6.6 x 10 M 0.10 M 4.4x10 10 M (c) From the definition of molarity, the initial quantity of H and can e determined as follows: H 0.10 M 0.10 M L L # moles of solute 5.0 x 10 4 moles # moles of solute 2.5 x 10 3 moles The hydrogen ion from the acid will react to aniline as follows: H (aq)c 6 H 5 NH 2 (aq) C 6 H 5 NH 3 (l) The reaction will proceed to completion until the limiting reagent is used up. Since there is less hydrogen ion to egin with and it reacts in a 1:1 ratio with the aniline according to the alanced equation, the hydrogen ion is the limiting reagent. The following tale summarizes the stoichiometry of the reaction: C 6 H 5 NH 2 H Initial 2.5 x x Change 5.0 x x x 10 4 Equilirium 2.0 x x10 4 With the addition of the 5.0 ml of HCl, the total volume of the solution is 30.0 ml ( ). So the final concentrations of aniline and its conjugate acid (C 6 H 5 NH 3 ) can e calculated as follows: C 6 H 5 NH x 10 moles C H NH L M C 6 H 5 NH x 10 moles C H NH L M x 10 M OH C 6H 5NH 3 poh log[oh ] log(1.7 x 10 9 ) 8.77 OH M M So [OH ]1.7x10 9 M. ph poh ph

2 1. (continued) (d) The equivalence point is the point in the titration where enough hydrogen ion (acid) has een added to completely react with all of the ase. Since, there was 2.5 x 10 3 M of C 6 H 5 NH 2 to efore the titration and the mole ratio of aniline to hydrogen ion is 1:1 then the equivalence point occurs when 2.5 x 10 3 M of acid is added. The quantities of the participants of the reaction is summarized in the following tale: C 6 H 5 NH 2 H Initial 2.5 x x Change 2.5 x x x 10 3 Equilirium x 10 3 AP Chemistry 2003 The volume of acid added to accumulate 2.5 x 10 3 M of H can e calculated from the definition of molarity as shown elow. This volume is added to the original volume of aniline to get the total volume of the solution after the titration. This total volume is used to calculate the concentration of at equilirium as shown elow on the right. Since C 6 H 5 NH 3, a weak acid, is the only major species at equilirium, the following reaction illustrates the dissociation of this acid aq H aq aq The a for this reaction can e found y dividing the w for water y the for the reaction shown aove 1. (a) as follows: 14 w 1.0 x x 10 5 a. Since C 10 6 H 5 NH 3 dissociates 4.4 x 10 equally into H and C 6 H 5 NH 2,[H ][C 6 H 5 NH 2 ]x. H C6H5NH2 a x x 0.05 x[h ]1.1x10 3 M ph log[h ] log(1.1 x 10 3 ) ph M H x 10 moles of acid L of acid Lofacid2.5x10 2 L25mL x 10 moles C H NH L 0.05 M (e) Erythrosine would e the most suitale indicator for this reaction ecause it has a pa (3) value that is closest to the ph (2.96) of the reaction at the equivalence point.

3 1. (a) ph log[h ] 4.35 log[h ] AP Chemistry 2002 [H ]1.1x10 5 () a H OBr x M x x 10 (c) (i) The equivalence point will e reached when the numer of moles of OH added from the Ba(OH) 2 equals the numer of moles of hyporomous acid efore titration egins. numer of moles of solute Volume (L) of solution numer of moles M numer of moles of moles OH 3 1 mole Ba(OH) 2 1 L of Ba(OH) 2 solution? ml Ba(OH) x 10 moles OH 2 moles OH moles Ba(OH) 2 Volume of Ba(OH) 2 solution L of Ba(OH) 2 solution 41 ml of Ba(OH) 2 solution (ii) Greater than 7 ecause the titration of a weak acid with a strong ase produces a asic solution at the equivalence point. OBr H 2 O OH 14 w 1.0 x x x x x 10 OH OBr x x a x [OH ]6.2x10 4 M ph poh (log[oh ]) log[6.2 x 10 4 M] ph > 7 H OBr 9 (d) a 2.3 x x x 10 OBr Note: OBr [OBr ] M [NaOBr] that must e dissolved numer of moles of OBr total Volume x L 9.5 x L L M (e) There are less oxygen atoms, which are very electronegative, in than HBrO 3. This means that the electrons are not pulled from the Br atom as muchin than HBrO 3 making the HBr ond less ionic and harder to ionize in water.

4 AP Chemistry 2001 massofsolute 3. (a) % mass x 100% g x 100% massofsolution 200. g % mass 16.3% ()?gh1.20 g H 2 O 200. gh 180. gh 2O g H PV nrt 750mmHg (3.72 L) n CO2 ( Latm/mol)(298 ) 760mmHg n CO mole CO gc?gc0.150 mole CO 2 1moleCO 1.80gCO 2 2 m C m H m O m Total 1.80g 0.133gm O 3.00 g m O 1.07 g m C g m H 1.80 g m O 1.07 g (c) molarity # molesofsolute Lofsolution # molesofsolute M NaOH Lofsolution. # moles of solute moles NaOH H OH H 2 0 Since the numer of moles of OH equals the numer of moles of H (the mole ratio is 1:1) during neutralization and there is only one ionizale hydrogen, then the numer of moles of NaOH equals the numer of moles of acetylsalicylic acid gacid? g acid 1 mole acid molesacid 1mole acetylsalicylic acid 180 g (d) molarity # molesofsolute Lofsolution x10 moles Lsolution M acetylsalicylic acid (i) HA H A ph log[h ] 2.22 log[h ] [H ]6.03x10 3 M[A ] [HA] M6.03x10 3 M M a [ ][ H A ] 3 3 (. 603x10 ) (. 603x10 ) 2.86x10 4 a [ HA] 0127.

5 AP Chemistry (a) H (aq) OH (aq) H 2 O(l) () See next page for calculations that descrie details of graph. These are unnecessary for this test since this section assumes no calculator and only a general form of the graph is necessary here. However, if you are interested, you can look to the next page ph Volume of 0.20 M HCl Added (ml) (c) Methyl Red is the est indicator ecause its p a (5.5) is indicative of an acidic region. The equivalence point from the titration of a weak ase with a strong acid will e in the acidic region (elow ph of 7). For mathematical details, see the next page (actual equivalence point is 5.26). (d) The resulting solution will e asic as shown elow. NH 4 NH 3 H w a x 5 fornh3 18. x10 NH a 5.6x H [NH 4 NH 4 ] [NH 3 ] Therefore a 5.6x10 10 [H ] This will give a ph in the asic region. Mathematical details: ph log([h ])(5.6x10 10 ) 9.25.

6 8. () Mathematical Explanations from previous page. Calculate initial ph (let x[oh ] at equilirium). NH 3 H 2 O NH 4 OH AP Chemistry 2000 Initial Change Equilrium [NH 3 ] [NH 4 ] [OH ] x x x x x 10 3 NH 4 OH NH 3 Before Titaration: 1.8x10 5 xx 010. x1.34x10 3 M [OH ] poh log [OH ] log(1.34 x 10 3 ) 2.87 ph poh Equivalence Point Determination: molarity molesofsolute Lofsolution molarity molesofsolute Lofsolution # molesnh M NH M HCl moleshcl Lsolution # Lsolution # moles NH moles NH 3 # L HCl solution L HCl solution Therefore, 15 ml HCl solution is the equivalence point of this titration. Equivalence Point Concentrations: NH 3 HCl NH 4 Cl This reaction goes to completion ecause ammonia reacts readily with a free proton. Therfore, the numer of moles of NH 4 the numer of moles HCl added ( moles) and the concentration of NH4 at the equivalence point (total volume is 40 ml 15 ml 55 ml) is: molarity molesofsolute Lofsolution molesnh Lsolution 5.5x102 MNH 4 The dominant equilirium at the equivalence point is given y: NH 4 NH3 H w a x 5 fornh3 18. x10 NH H a 5.6x NH 4 Since is small, x is small compared to 0.10 M, so [NH 3 ] is 0.10 M at equilirium. 5.6x10 10 xx letx[h ] 55. x10 2 x[h ]5.55x10 6 ph log(5.55 x 10 6 ) 5.26 Beyond the Equivalence Point: Since NH 4 is such a weak acid, the main source of H will e excess HCl (after equivalence) which is 25 ml of HCl solution. The total volume of the solution at the end is (40.0 ml 30.0 ml 70.0 ml). molarity molesofsolute Lofsolution molarity molesofsolute Lofsolution molesh Lsolution