Chem 222 #20 Ch 11 Nov 1, 2004

Transcription

1 Chem 222 #20 Ch 11 Nov 1, 2004

2 Announcement The answers for Midterm will be uploaded sometime tomorrow. Please check your grade and talk to your TA about general questions If you have specific questions about grading, talk to the following persons: Q1. Koy, Q2. Amanda, Q3. Chris, Q4. Medhat, Q5-6. Ishii Please complete correction by next Wednesday. We do not accept any corrections after that. Do not modify your midterm if you want us to correct your grade.

3 Histogram for Midterm Series < Average ~ Cf. Average of quiz ~ 68

4 Q5 1) Consider the titration of ml of M Ce 4+ in 1 M HClO 4 by M Fe 2+ to give Ce 3+ and Fe 3+, using Pt and calomel reference Hg Hg 2 Cl 2 electrodes to find the end point. E = Log([Fe 2+ ]/[Fe 3+ ]) (V) E = Log([Ce 3+ ]/[Ce 4+ ]) (V) a) Calculate E at the following volume (V Fe ) of Fe 2+, V Fe = 2 ml and 30.0 ml. a) First, obtain V Fe at the equivalence point (V e ). V e = C Ce4+ V Ce4+ /C Fe2+ = ml*0.0100m/ M =20.0 ml (2 points) When V = 2 ml, Use [Ce 3+ ]/[Ce 4+ ] and obtain the ratio. E = Log([Ce 3+ ]/[Ce 4+ ]) = Log{(V/(V-V e )) = Log{(2/18)} (2 point) Note: No points if the equation for F e is used or the concentration ratio is incorrect. E = V (Correct calculation 1 points) When V = 30 ml, Use [Fe 2+ ]/[Fe 3+ ] and obtain the ratio E = Log([Fe 2+ ]/[Fe 3+ ]) = Log((V-V e )/V e ) = Log((10.0 ml)/(20.0 ml) (2 points; No points if the equation for Ce is used) E = V (1 point)

5 How we calculate the titration? Ce 4+ + Fe 2+ Ce 3+ + Fe 3+ Titrant Analyte What must be changed? Before equivalence Excess Fe 2+ & Fe 3+ Little Ce 4+ After equivalence Excess Ce 4+ &Ce 3+ Little Fe 2+ Potential (V versus S. C. E)

6 Q5 1b) Consider the titration of ml of M Ce 4+ in 1 M HClO 4 by M Fe 2+ to give Ce 3+ and Fe 3+, using Pt and calomel reference Hg Hg 2 Cl 2 electrodes to find the end point. b) Calculate the concentration of Ce 3+ when V Fe = 10.0 ml. At V Fe = V e /2, [Ce 4+ ] = [Ce 3+ ]. So [Ce 3+ ] = /2 *(100.0/110.0) = mm

7 Q5. 2) Find the ph and concentration of (CH 3 )N and (CH 3 )NH + in a M solution of trimethylaminium chloride ((CH 3 )NH + Cl - ). The pk a of (CH 3 )NH + is You can neglect the effect of Cl - K a = 10 -pka = = (px = -logx X = 10 -px ) (2 points; No points if the definition of pka is incorrect) (CH 3 )NH + (CH 3 )N + H + F x x x F = M K a = x 2 /(F-x) ~ x 2 /F (assume F >> x) x = (K a F) 1/2 = ( ) 1/2 (3 points) = M ( 1 point) (F-x) ~ F ph = -logx = (2 points) Or ph = pka + log[ch 3 N]/[CH 3 NH + ] (2 points) (10-16) ph = pka - ph log(f-x) (ph = px = -logx) ph = {pka -log(f-x)}/2 (4 points) ~ (pka - logf)/2 = {9.800 log(0.080)} = X= 10 -ph = (2 point) If you just answer [(CH 3 )N ] = F, I will give 2 points.

8 Q6 Q6. Extra question Total 5 points g sample contained only NaCl and KBr. It was dissolved in water and required 2.00 L of M AgNO3 to precipitate all Cl- and Br-. The molecular weights of NaCl and KBr are 58.4 and 119.0, respectively. Calculate how many moles of KBr are contained in the solid sample. The question is from Home work Assume the contained KBr is x mmol. Ag- +Cl- AgCl Ag- +Br- AgBr Total Ag moles is 2.00 L M = 3.00 mmol (2 points) Total weight of NaCl and KBr is x mg (3.00 x) mg = mg (2 points) ( )x = x = 90.9/( ) = 1.50 mmol (1 point) No points will be given to 3.00 mmol/ 2 =1.5 mmol. The assumption that the AgCl and AgBr are equal in the number of moles is incorrect.

9 Mixing a weak acid and its conjugated base (slide 4 on Oct 21) HA H + +A - pk a = 4.00 K a = 10 [Q1] F- x x x F = 0.01 M x 2 /(F-x) = K a x = Fraction of dissociation α =[A - ]/{[A - ] +[HA]} = x/f = =3.1 % Remember Le Chatelier s principle? (p103) What happens if you add A - to the solution HA H + +A - How much fraction of A - reacts with water in a solution containing 0.10 M of A -? A - +H 2 O HA + OH - pk b =[Q2] F y y y y 2 /(F -y) = K b y = & α =

10 P206 Compositions of 0.05 M Solution The acidic form H 2 L + Leucine hydrochloride contains protonated species. Because K 1 = , H 2 L + is a weak acid. HL is an even weaker acid: k 2 = It appears that the H 2 L + will dissociate only partly, and the resulting HL will hardly dissociate. H 2 L + HL + H + F x x x K 1 = [Q1]/(F x) x= M K a2 = [H + ][L - ]/[HL] Basic Form L - The species, L -, is found in a salt such as sodium leucinate. In solution, the salt gives L -. L - +H 2 O HL + OH - K b1 = K w /K a2 = 5.59 x 10-5 HL +H 2 O H 2 L + + OH - K b2 = K w /K a1 = 2.13 x K b1 shows that L - will not hydrolyze very much to give HL. K b2 shows H 2 L + is generated even less. L - +H 2 O HL + OH - K b1 = [Q1] F-x x x K b2 = [H 2 L + ][OH - ]/[HL]

11 11-1 Diprotic Acids and Bases H 3 N + - O 2 C \ CH-R Substituent / where R is a different group for each amino acid. ph = pka + Log[A - ]/[AH] [A - ]/[AH] = 10 (ph-pka)

12 Unusual serine sidechain in serine protease (trypsin) Catalytic triad in trypsin (Ser195)-O-H + His (Ser)-O - + H-His Serine protease: subtilisin, chymotrypsin,

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14 Intermediate Form, HL A solution prepared from leucine, HL, is more complicated. HL H + +L - ; K a2 = (11-8) HL + H 2 O H 2 L + +OH - ; K b2 = (11-9) A molecule that can both donate and accept a proton is said to be amphiprotic. To solve (11-8) & (11-9), we use charge [Q1] [H + ] +[H 2 L + ] = [L - ] +[OH - ] [H 2 L + ] = [HL][H + ]/K a1 [L - ] = [HL]K a2 /[H + ] [OH - ] = K w /[H + ] [H + ] + [HL][H + ]/K a1 = [HL]K a2 /[H + ] + K w /[H + ] {1+([HL]/K a1 )}[H + ] 2 = ([HL]K a2 +K w ) Major species is [HL] [HL] ~ F [H + ] = {(FK a2 +K w )/{1+(F/K a1 )}} 1/2 = 1/ 2 (11-11) FK a2 K K a1 a1 + K + F a1 K w

15 [H + ] = FK a2 K K a1 a1 + K + F a1 K w 1/ 2 Two additional assumptions K a2 F >> K w [H + ] = FK K a2 a1 K +F a1 1/ 2 K a1 << F [H + ] = FKa2K F a1 1/ 2 = ( K K ) 1/ 2 a1 ph = -Log[H + ] = -Log(K a1 K a2 ) 1/2 = {-LogK a1 LogK a2 }/2 = [Q1] (11-12) a2

16 Ex. ph of the Intermediate Form of a Diprotic Acid (p210) KPH is a salt of the intermediate form of phathalic acid. Calculate the ph of both 0.10 M and M KPH. F >> K 1 = 10 [Q1] & FK 2 >> K W = ph = [Q2] = 4.18

17 Intermediate Form, HA - Summary A solution prepared from the intermediate A a1 form HA -, is more complicated. HA - H + +A 2- K a2 (11-8) HA - + H + H 2 ; 1/K (11-9) ph = (pk a1 + pk a2 )/2 [H + ] = 10 -ph [HA - ] = F K a2 = [H + ][A 2- ]/[HA - ] [A 2- ] = K a2 [HA - ]/[H + ] 1/K a1 = [H 2 A]/[HA - ][H + ] [H 2 A] = [H + ][HA - ]/K a1

18 Read the summary p211

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21 11-2 Diprotic Buffers H 2 A HA - A 2- You can use either or both of the equations ph = pk 1 + Log[HA - ]/[H 2 A] ph = pk 2 + Log[A 2- ]/[HA - ] Ex. Find the ph of a solution prepared by dissolving 1.0 mmol of KHP and 2.0 mmol of Na 2 P in a 1L of H 2 O. ph = pk 2 + Log{[P 2- ]/[HP - ]} [P 2- ]/[HP] =

22 Ex2. How much volume of M NaOH should be added to 0.80 L of 0.50 M KHP solution to give a solution at ph of 6.4. pk 1 = 3.0 & pk 2 = 5.4 Initial moles of KHP is 0.80 L 0.50 M= 0.40moles HP - + OH - P 2- +H 2 O Initial 0.40 x Final 0.40-x ph = pk 2 + Log[P 2- ]/[HP] [P 2- ]/[HP] = 10 {ph-pk 2 } = x