122ch17c - The volume of NaOH so‘lution required in all...

Info iconThis preview shows pages 1–9. 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

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

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The volume of NaOH so‘lution required in all cases is ' vm x Mm g (0.100) vac v = “1 =1.25v “m (0.080) ‘°‘°' 1 Mass The coricentrafion of the salt at the equivalence point is 0.100 2.25 Maud Vacid g = 00444 M 2.25 vm _0) (0 000444 M’NaBr. pH = 7.00 .51. 0.0991/ -7c Kb: [HCaHsoalIOH‘] a _K_w_ = mono-14 g [03H503] [Hoar-1503] = [OI-1‘]; [03H503'] 2 0.0444 [OH‘]2 = 004440.14 x 10'11 ); [OH'] = 1.78 x 10'6 = 1.8 x 10'6 M [13”;03'}Kb Cr04'2(aq) + H200) ‘——'r HCrO4'(aq) + OH’(aq) amwv¢ 76 7‘ ' Kb = W: «Kl = mmfi1 (0) "If _ Kn [on-r12 = 0.0444(3.33 x 10-5); {or-r] = 3.845 x 10-5 = 3.8 x 10-5, [gran/3'] Kb H1H|H|HV||HIH|III . The total voiume at the equivalence point is me + Vm = 2.25 Vac”. ' {A 525' H) W 0.0444 MNa‘CaHsoa“: 03H503‘(aq) + H200) "—r" HC3H503(aq) + 7C . 7.14 x 10‘" =7.1 x 10'11 D pOH = 5.75; pH = 8.25 pal/=17: 9/, /a//= 9,5? Cay—(3CD Z n {9, r / v y!!! Wer 7 . The equilibrium of Interest ls ' L— . _ . = x _‘ = [H ——~____ H05H303(aq) .__, H (aq) + 051-1303 (aq), Ka 6.76 10 mchsHaoal —' Begin by calculating [HC5H303] and [051130;] for each case. 35.0 g Hcfii-Izp3 x 1 mol H05H303 0.250 L-soln 112.1 9 HC~,,H303 30.0 g Nat‘;5H303 x 1 mol NacsHa'o3 I 0.250 L soln 134.1 g Nacstp3 m g KJHCsHsosl = 6.76 x10*g1.249 -a¢ : 6.76 x10'4(1.249) [canaoz’] (0.8949 +x) _ (0.8949) (6) = 1.249 = 1.25 M H05H3O3 = 0.8949 = 0.095 M 0511303- [I-l‘].= 9.43 x10-4 M, pH = 3.025 ' 3'- 9062; ' (b) For dilution, MN1 = MZV2 [HcsHaoa] = 0.250 M x 30.0 mL m ...... .. 125 m = 0.0600 M [C5H303'] = W = 0.0352 M .4 [H‘] 2 W = 1.15 x10“3 M. pH = 2.938 {435 7' %’M 76 )5 fm// ¢Jm< 1.er /' m — /,{_/1 me ~ /'5 a 45L. perm”: o o 426/ fl“) 0.0850 M x 0.500 L - 0.0425 mol H05H303 1.55 M x 0.0500 L =0.0825 mol NaOH _ HC5H303(aq) + NaOH(aq) -) NaC5H303(aq) + H200) initial 0.0425 moi 0.0825 moi ' reaction -0.0425 moi -0.0425mol +0.0425 moi __ after 0 incl _ ' 0.0400 moi 0.0425 moi The strong base NaOH dominates the pH; the contribution of 05H303' is negiiglble. This combination wouid be “after the equivalence point” of a titration. The 'tbtal volume is 0550 L. [OH] = 0.0400 mo! 0.550 L = 0.0727 M; pOH = 1.138, pH = 12.862 a 7’7 7 7 " ’ “5'7x "’ ' (a) HA(aq) + B<aq> : HB*(aq) + A-<aq> K; = w ,> /v"- 5’- 5’ “‘— IHA] [B] ' (b) Note that the solution is slightty basic because B is a stronger base than HA is an acid. (Or. equivalentiy, that A‘ is a stronger base than HB” is an acid.) Thus. a little of the A‘ is used up in reaction: A-(aq) + H200) é; HA(aq') + OH”(aq). Since pH is not very far from neutral, it is reasonable to assume that the reaction in part (a) has gone far to the right, and that [A‘] = [HB‘] and [HA] z [B]. Then —" I _K,= M51 = 5.0 x10-6; when pH =83, [H"]== 1.58 x 10-9= 1.6 x10'9M W _.___ [HA] __ ————— i’fl=5.0x10'°/1.sax10'9=3.155x1o3=3.2x103 . .— . [HA] IQ [4t] 2 _ m. _.___ From the assumptions above, m- = m, so :- [A1 = 9.953 x 10“ = 1.0 x 107 — . [HA] [3] {HAP I X S 2E 77 M_, initial molHA= “355 $1.334):10-=*=1.i33x10*3 mol HA 73.9 g/mol ‘ mol OH“ added to pH 5.10 = 0.0950 M x 0.0120 L = 1.14 x 10'3 mol OH' HA(aq) + NaOH(aq) -) NaA(aq) + H20 ' before m 1.834 x 10-3 mol 1.14 x 10'3 mol 0 change -1.14 x10'3 moi -1.14 x 10‘3 mo! +1.14 X10'3 mol after m 0.694 x 10‘3 mol 0 1.14 x 10'3 mol 0.694 x 10* mol [HA] - _____. 0.01874 = 0.0187 M 0.0370 L g g 1.14 x10'3mol A. l 1 0.0370 L = 0.03081 = 0.0308 M; [H‘] = 10‘5-‘° = 7.94 x 10‘5 M - The mixture after reaction (a buffer) can be described by the acid dissociation equilibiium HA(aq) ‘——r H*(aq) + A'(aq) initial 0.0187 M 0 0.0308 M equil (0.0187 - 7.94 x 10‘6 M) 7.94 x 10'6 M (0.0308 + 7.94 x 10'?) M. --— K‘é [i-lj[A‘] z (7.94x10'°)(0.0308) 31,3,‘104 '_ the data indicate an answer with 2 significant figures.) _ [HA] (0.0187) (Although we have carried 3 figures through the calculation to avoid rounding errors. ‘ ’- I, - I Ill /', 73“ I 4/ '4 92/7‘ // ’ 0 ’ y ,r/ ’ ¥ 7" ‘2 v 7'." __,_(a) K,.=%1; [H‘]=1O"‘°°=1.58#10‘5é15x10‘5M Assuming [H‘] is small compared to [HA] and [A‘], K3 2 (1.53 X10'5)(0.080) =1.06 x10“; =1‘1x10'5. = 4.98 (0.12) . pK' II | (b) At pH = 5.00, [H‘] = 1.0 x 10‘5 M. Let b = extra moles NaOH. ——~-—— [HA] = 0.12 - b, [A‘] = 0.080 + b 0% ‘0 ’4) ——~-~ IQ: 1,06 x 10-5 : W; 2.06 x10'5 {3: 4,72 x10”; _________ ._.._.—_.___.._ w ~—-—~——-—- b = 0.023 moi NaOH __—'C3H503- be by reaction 0f Heal-I503 With ’ 0.1000 M x 0.05000 L = 5.000 x 10‘3 mol HC3H503; b = mol NaOH needed 0.4427 (5.000 x 10—3 - b) = b, 2.214 x 10-3 = 1.4427 b, b (5.000 x {0'3 — b) = 1.53 “10'3 Heal-1503 + NaOH -) C3H503‘ + H20 + Na” initial 5.000 x 10‘3 b mol - 1x -b mol -b mol +b moi after rx 5.000 x 10‘5 - b mol 0 b mol C H O - _ K,= M; .K, = 1.4 x104; [H‘] =10-pH =10-3-5o = 3.16 x104 = 3.2 x10'4M __ [HcaHsos] _— Since solution volume is the same for P10311503 and 03H503‘. we can use moles in the - ' equation for [H‘]. —_ A4 - . ' “—K.=1.4x1o-4=—...—.3-16"1° 0’) . —. (Using 1.53 x 10'3 — check.) — Calculate volume NaOH required from M = mol/L. 1.53 x 10'3 mol OH‘ x Substituting this result into the K, expression gives [H‘] = 3.27 x 10‘4 . = 1.5 x1043 mol OH' (The precision of K8 dictates that the result has 2 sig figs.) mol OH‘ (3 sig figs) gives, [H‘] = 3.16 x 10", a more reassuring cross 1L 1 3.1L 1.000 mol 1X10‘eL X =1.SX103uL(1.5mL) 1mol Cat-170260014 x tablet 1000 mg 180.29 Cal-1702000?! 1L caH702c00H(aq) ‘——7 C3H70'2C00'(aq) + H+(aq) initial 3.61 x10-3 M o 1 x10'2M 325 "‘9 x 2tablets x A. x J|||I equil (3.61 x 10-34) M xM (1 x10‘2 +30 M — x= [c3H702000'] = 1.08 x10'5=1 x10's M .08 '5 COO ' % ionization = 1- x 10 M CBH702 —_.._....__ x 100 = 0.3% — 3.61 x 10'3 M CBH702000H (% ionizatiOn is small. so the assumption was vaiid.) % aspirin molecules = 100.0% - 0.3% = 99.7% molecules ...
View Full Document

This note was uploaded on 07/17/2008 for the course CHEM 122 taught by Professor Zellmer during the Summer '07 term at Ohio State.

Page1 / 9

122ch17c - The volume of NaOH so‘lution required in all...

This preview shows document pages 1 - 9. Sign up to view the full document.

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