Chemistry 271 - 23XX

Prof. Jason Kahn

University of Maryland, College Park

General Chemistry and Energetics

Exam II (100 points)

You have 53 ·minutes for this exam.

Your Name:

Your SID#:

Your Section # or time:

April 6, 2016

Explanations should be concise and clear. There is extra space on the last page if you need it.

You will need a calculator for this exam. No other study aids or materials are permitted.

Generous partial credit will be given, i.e., if you don't know, guess.

Useful Equations:

Ka= [H+][A-]/[HA] pH= -log([H+])

R = 0.08206 L·atm/mole K k8 = 1.38 x 10-23 J/K

R = 8.314 J/mole K = 1.987 cal/mole K = NAkB °C = °K-273.15 P(v)dv 7 Cv2exp(-mv2/2kT)

pH= pKa + log([A-]/[HA]) KP= K/RT)&i

Kb = [BH+][HO-]/[B]

lnKeq = -Af-l°/(R7) + M0/R

KaxKb =Kw

Ink= (-E/RT) + In A

Kw= [H+][oH-] = 10-14

Honor Pledge: At the end of the examination time, please write out the following sentence and sign it, or talk to me about it:

"I pledge on my honor that I have not given or received any unauthorized assistance on this examination."

Chemistry 271, Exam II, 23xx, 4/6/2016

1. Straightforward Acid-Base {30 pts)The equation for the dissociation of benzoic acid in water is

PhCOOH � H+ + PhCoo-

2/9

Ka

= 6.25 X 10-5

(a; 3 pts) Calculate the pKa of benzoic acid.

@ P.\ie. ::: - laJ ( [4- � ccc -If ( b. z5ic(v'') =- �, 2,011 @

� [ +L 1,/z,i j.,-, 4,Lr '1. lo .,.,J."<J� LtJs,,;}(b; 9 pts) Calculate the pH of a 0.150 M solution of benzoic acid in water, assuming x << C0 • Explain or

demonstrate why the x << C0 condition is not satisfied for a 0.000015 M solution of benzoic acid.

:J?- C ft, }t. l Lt '.

I

C

fi

Ph coo vt

6. 19) W\

-x

Dfg>- X

p� c0u6 vi +

-t> ·o

+x t)C

'X

c;._ f h (Ol)v1 I � /.. {DI) - � v.P.,;.J,J {,,_ � evf/ L d ef j;. }.. ( .Y :!, I,�) w- :b I) p l1 '-'v1AJ 1:,.0 fv,,,,,. 1k f )!_,

) w

�h Cl;� �·(!_[A Z,� - fVI �2 {oJ �t-s"'.� w"3,2, r-S',2J

D l.JJ1� .µ � � u11-or [f�CtrtJvtJ CJ/' [1� Cc-or] IS ye) -

,;.,,..J.)_ -1i4- ,i, ,,i' ./1 C,� wwl� ,,,p,.,.- ,,)/,1-,"" f tuA?I. Pv

b ,-s,<_ - � sv-�1- � /1.::t 11\0 '1-H � f;,c,' 6.J- N+ "': Score for the page __-4---LI�

Chemistry 271. Exam II. 23xx, 4/6/2016 3/9

(d; 6 pts) Write down the base dissociation equilibrium reaction and the equilibrium constant expression for benzoate (PhCOo-, = the conjugate base of benzoic acid). Calculate the pKb for benzoate.

[f h Coo-�] L );to�[fk coo8]

flL� = IY- f 1, = /Lf.-4.2o'f @

z � e r -:;-c;b--

(e; 8 pts) Calculate the pH of 0.100 M NlPhCoo- (the pH you would get by dissolving 0.1 moles of sodium benzoate in 1 L of water). How did you know before you started that the answer must be > 7?

[ us-a� c,.-,nf..A':J l(1 � �. :S hi,..]

[ ft., (o-v�J f 11-t] --, -

[ft._ (Oo�J

� f [Ir =- � I Lj - p O IA = 1 '1 - ��. 3 "iJ" = 8, 6 D 2.. @

�it� ��� l] � �

@ (No.On\

'l Q, cft?A. { PL ·Co-uvt � �

f AJ,c fl (()o-

Score for the page __ -4-/_I_L_J

CbemisID Tl. faam II 23xx 4/6/2016 4/9

2. Liukqe and Equilibria {30 pts)(a; 10 pts) ·we described in class that protonation of hemoglobin is linked to the interconversion of Rand T

states. The R state binds 02 much more tightly than the T state. What acid is generated by aerobic metabolism? Which state of hemoglobin is favored at acidic pH? Why is this important for mammalian physiology? - fl--'(;-

:,� l- ��,.� � CQ'l- 1 w�'u/r d�-1i,lve,,--�

�v< &�2. .,;��

,-ii,

(e,3

, (@¢'ra?) , [!Af/flv eu.,'X : <fDJ - 1--, -r- S-1

(.t. .1 /:._J,y,) 1\,-, cu,'/ rt- c.-Jl9,"f! ( tla Bohr�).

@ - --,k � li..f o..divrJ;; <.Jwl,';J � or /-,?fv<4.P- l,,r,.rG '

.. l ,v ,I} H rv,J.. J ,� hh Iv si,,,'rr 1i,.h h @u-\.{) r J - -

T .<�It) J.,1L- � tt& tt> cR�tit/L�1- '"' k

-h'3.<� f'2

The DNA double helix can bend to form small circles (hundreds of base pairs), and the circles are readily distinguishable from linear DNA. The diagram below illustrates equilibria between linear DNA on the left and circular DNA on the right. Transcription factors are proteins that bind to DNA, and in some cases they induce DNA bending, as shown for the double-quadrilateral protein. The equilibria for protein binding to linear or circular DNA are shown running vertically, with free DNA at the top and bound at the bottom. Finally, the bold lines indicate a stretch of DNA that is permanently, intrinsically curved.

Linear DNA

P�t�-J rKbind,linearroem J

Bent DNA

'

Kc.bound ' '

Score for the page __ �/,_

.1-1,lC.,£.2-

Chemisu:y 271. Exam II, 23xx. 4/6/2016

(b; 8 pts) It has been found experimentally for some systems that Kc.bound is 200x larger than Kctree· Write

down the mathematical relationship among the four equiJibrium constants in the diagram. What is the

value of Kbind,circte I Kbind,linear? Explain the linkage in words.

5/9

Chemistry 271, Exam II, 23xx, 4/6/2016

3. Titrations and Amino Acids (40 pts)

Acid-Base Titrations 14

,----��.-��-.-��---r���.---��.-----��--r-��--.-��---,

I pKb = o I 12

10

I pKb = 4 I 8

I pKb = 6 I 6

I pKb = 8 I

Added volume of Base

(a; 9 pts) The graphs above show five titrations of the same weak acid, at the same concentration, using

bases with pKb's of 0, 2, 4, 6, and 8, all at the same concentration. Explain why the five curves all

coincide on the left and diverge on the right. Estimate the pKa of the weak acid.

6/9

Chemistry 271, Exam II, 23xx, 4/6/2016 7/9

(b; 9 pts) Consider the right hand edge of the graph, where 2 equivalents of base have been added. What is the ratio of [BH+]/[B] at this point for titration with a weak base? Derive the relationship apparent from the graph that pH= 14- pKb at this point, for weak bases. In class we sometimes described the titration curve after the equivalence point as "adding base to salt." This is accurate only for titration with a strong base, for which the final pH is the same as the pH of 100 ml of base diluted to 200 ml. Qualitatively explain why is this not true as for weak bases: the pH at the end is less than the pH of pure diluted base.

,,,- \{c..lf � � � � � J1\tr.,, /J,-"'

,r � ),t +-, � of�r lft-1)

@ vwi� ,<f-\'1, v2, ¥ c. =l k1r,, 5,£) l�tvi 1J /U.J)= I

-- ·1.(2 we Lk'--J tlu.r- It.- 11 {1 ,A; ,r � cu;:J [ !3 V1 f-J)

{.,lit �

@ t* "' f' K'.0-. + (� [ ttkthiJ "- f kc.. ) bJ -11--< tf /{� -=- I 'I -p }'!,

bf 11 � JLj" f j{\, v" {+3) �

I-� -- � �ro--lL h�Jls

J Ji_ .,fr[( [ glA tJ 5tA/J rffS� � �� IA,/ }1 � 1 i pH K;

Sf (e; 10 pts) The structure of fully-protonate arginine is shown below. Draw the predominant ionic form at J �..}­pH 10.75. What is the special name/symbol we give to this particular pH, for Arg? Why is the pKa for � 4. h_the guanidino group at the left so much higher than the.pKa for the protonated amine group? b J

pKa = 12.5 pKa = 1.8 �

c;NH2 0

�H 2N_ if · N l:_; H

-

OH

,) e, � w-erlt �.?.,,{ � tu @ ���

_ (! � Score for the page __ ......,/,_+-f l-+--Y'C�., �

Chemisgy 271. Exam II. 23:u. 4/6/2016 8/9

(c; 12 pts) Sketch a titration curve for titrating 100 ml of 0.1 M sulfurous acid, H2S03 with 0.1 M NaOH.

The first pKa of H2S03 is 1.77 and the second is 7 .21. The initial pH is 1.34. Write in the pH values at

the three indicated volumes of NaOH.

12

10

8

:I: C.

6

4

2

0 0 25

pH= /,11"' -; (I lli I

50 75

Acid-Base Titration

100 125

·Added volume of Base

V L VfJv.'V

( J._11 ;/-,! I ::. �·:: �. �lL/ 2-

I

\�

Z, w{ �·y-1- (] c$ [1;o'J Z � .c:

/O,\Q<

�(/._.

f,f• +-

@v Cvt�1�

i a_l'r</�

� 25!" *' I ::. 1 o -{.Y

150 175 2 0

J:1, .. d1w--f CJ,\�

I :: O.D( �

2. 1-'t•v (Vl n ... � V i,t,v

� ·::

V 1(5· , .. I

1 -Jgi:::: 0

put1] lPt

Score for the page. __ _..;./_l 2-_

= -- 9/9

Page Score

2 /� 3 � 4 a.z-.---5 jl-8"'

6 � 7 /� 8 n.o-- I 2--

. ..---Total /100 \ ()O ./

Score for the page ___ _