### BIOCHEM study questions

greenspun.com : LUSENET : NJMed Class of 2003 : One Thread

These are M. Angelucci's answers. I started a new thread for them.

1. =[OH-]

2. >[OH-]

3. pH = -log[0.5 EE -3] = 3.3

4. 4.9 = 4.7 + log x/(0.06) x = 0.095M

5. 4.9 = 4.7 + log x/(0.1) x = 0.16M

6. 9.5 = 9.8 + log x/(0.1) x = 0.05M

7. 2.6 = 2.3 + log x/(0.1) x = 0.20M

8. (2.3 + 9.8)/2 = 6.05

9. (4.3 + 2.3)/2 = 3.3

10. a. -log[5 EE -4] = 3.3 b. 14 + log[7 EE -5] = 9.8 c. -log[2 EE -6] = 5.7 d. 14 + log[3 EE -2] = 12.5

11. a. 1 EE -2.9 = 1.26 EE -3 b. 1 EE -6.9 = 1.26 EE -7 c. 1 EE -9.5 = 3.16 EE -10

12. 4.8 = pK + log[0.087]/[0.01] pK = 3.86

13. 5.8 = 4.76 + log x x = 10.69

14. 7.4 = 3.8 + log x x = 4.0 EE 3

given [H+][HCO3]/[H2CO3]; remove H+ by transporting out of system, equilibrium will drive equation [H2CO3] ---> [HCO3] resulting in large ratio

-- Michael Angelucci (angelumi@umdnj.edu), August 24, 1999.

-- Omar Akhtar (akhtarom@umdnj.edu), August 24, 1999

Numerical answers for #1 and #2:

1. 7=-log(h+), so (H+)=10e-7

2. 2.5=-log(H+), so (H+)=10e-2.5 The rest of my answers were the same as above.

-- Omar (akhtarom@umdnj.edu), August 24, 1999.

Michael- I'm not sure about 5,6,7, or 13. For 5,6,7 I think the ratio of conjb/acid should be [x]/[amount of buffer added - x]. That'll give .061, .066, .066 for 5,6,7. 13 may've been a typo. Let me know whether I am mistaken. Thanks for posting your answers-

-- Matthew Trovato (trovatma@umdnj.edu), August 24, 1999.

-- Matthew Trovato (trovatma@umdnj.edu), August 24, 1999.

Matthew is right about the ratio of conjbase/asid -- I checked with Dr. Wagner today.

-- Irina Sigal (sigalir@umdnj.edu), August 24, 1999.

Thanks guys, I remember now and it's all coming back to me. I forgot about that 1-x thing goin on in the denominator, but good lookin out--->MikeA

-- Mike Angelucci (angelumi@umdnj.edu), August 24, 1999.

I'm not sure if I made a mistake in my calculation...but I got a ratio of 3.5/1 for problem #13. Did anyone else get this answer too? Also, I think that the reason that the ratio is so large for #14 might be that the carbonic acid serves as a temporary intermediate in the buffer system. In other words, when the H+ conc is high, the carbonic acid would rapidly convert to CO2 & H20. This CO2 could then be exhaled in the lungs.

-- Allison Froehlich (froehlal@UMDNJ.edu), August 26, 1999.

Allison, I rechecked my #13 and I also got 3.5 to 1.

-- Omar A. (foo@bar.com), August 26, 1999.

I got 0.033M as my answer to #6. inv log(9.5-9.8)= x/(0.1-x) 0.5(0.1-x)= x 0.05=1.5x x=0.033

How did someone get 0.066M for this? Thanks for posting answers :)

sancheru@umdnj.edu

-- Ruth Sanchez (sancheru@umdnj.edu), August 28, 1999.

Ruth, I agree with you on that question #6... I got 0.033 as well

But a general question to everyone... now that we all educated everyone on the x/(conc of buffer - x) and that the original question 5,6,7 were wrong... isn't question #4's answer wrong too then? it shouldn't be .095, but instead....0.037

4.9 = 4.7 + log (x / (.06 - x)) 10^.2 = x/(.06-x) .095 - 1.58x = x .095 = 2.58x .037 = x

did everyone else get this too?

-- Albert Li (lial@umdnj.edu), August 28, 1999.

Albert- Yes, you are right about #5 too. For any weak acid or weak base (i.e. it has a dissociation constant), you have to say pH=pK + logx/(HA or HB-x) where x=[conjugate acid or base]. So: #4 is 0.037 #5 is 0.061 #6 is 0.033 and #7 is 0.066

-- Ruth Sanchez (sancheru@umdnj.edu), August 28, 1999.

Regarding the current line of reasoning on #4, I have to disagree. In #4, the acetic acid (HA) concentration is *given* as 0.06 -- there is no need to subtract the x. I support this argument with a statement I learned from my undergrad biochem prof., one that has always steered me straight with buffer problems: "If the pH is less than the pKa, the protinated form predominates!" So, in this case, the answer of 0.095 is intuitively correct -- since the pH in the problem is *greater* than the pKa, the *non-protinated* form (in this case, acetate ion) should be predominating. 0.095 M acetate > 0.06 M acetic acid. BTW, just as an aside, there is only one significant figure in this problem, so the answer should be 0.1 M

-- Dave Ferrand (ferranda@umdnj.edu), August 28, 1999.

I think Dave's right about #4. The concentration of the acetic acid in the acetate buffer is 0.06 M, so there you don't have to follow the steps that were needed for #5,6,7. The answer should be 0.095 M or, more accurately to one sig-fig, 0.1 M.

For #6, the concentration of the zwitterion (+NH3CH2COO-) should be 0.067 M (or to one significant figure, 0.07 M) rather than 0.033 M. This makes sense because the pH is 9.5, so the more protonated species should predominate.

x=[NH2CH2COO-] .1-x = [+NH3CH2COO-]

9.5 = 9.8 + log (x/(.1-x)) x=0.033 M = [NH2CH2COO-] .1-x=0.067 or 0.07 M = [+NH3CH2COO-]

Has anyone done the study questions on p.22 of the biochem syllabus?

-- Anoop Nambiar (nambiaan@umdnj.edu), August 28, 1999.