# Biochem Problems

1. Sep 15, 2004

### RedVelvetCake

I need help with these three problems :

(1) Describe the preparation of 2.00L of .100M glycine buffer, ph 9.0 from glycine ) molecular wt in the zwitterionic form- 75.07g/mol) and 1.00M Naoh. tHe pka of glycine is 9.6.

(2) What is the ph of the following buffer mixtures ?
a. 1M acetic acid plus 0.5M sodium acetate
b. 0.3M phosphoric acid plus 0.8M KH2PO4.

(3) The weak acid HA is 2% ionized (dissociated) in 0.20M solution.
(a) what is the Ka for this acid ?
(b) what is the ph of this solution ?

Ive been trying to solve these problems for 2 days PLEASE SOMEONE HELP
Nikki in NY

2. Sep 15, 2004

### cronxeh

3. Sep 17, 2004

### so-crates

Those problems don't seem that hard. Just the use Henderson-Hasselback equation for buffers.

4. Sep 18, 2004

### chem_tr

pH calculations

Hello,

Glycine is converted to glycinate with the action of sodium hydroxide; the zwitterionic form is not valid at this point. The amino group is in the neutral form, NH2 and carboxylic group is COONa in this pH value.

In the second question, I think we should consider volumes equal to each other, as intended by the term "buffer solution". So the first one should be acidic, and and the second one is slightly basic.

In the third question, you'll need to write the dissociation equation of HA (0,20-x) --> H+ (x) + A- (x) and then, x/(0,20-x)=0,02.

If the acid or base ionizes less than 10%, then you may omit the x in (0,20-x) as it would make your calculations easier.

You may even try to solve a polynome by the discriminant equation, delta=b2-4ac, where your polynome is 0,20x-x2-0,02=0, equal to x2-0,20x+0,02=0, and a=1, b=-0,20, and c=0,02.

Regards, chem_tr

5. Sep 18, 2004

### RedVelvetCake

Thank you chem_tr for your help and so-crates mind your business if you dont have any positive feedback to give !!!!!!!!!!!!!!!!!!!

6. Sep 18, 2004

### GCT

I've replied to your other questions at chemistryhomework.com (replied to your post there).

Here's the answer for the first problem,

A buffer solution consists of a correlary ratio of concentrations of the acid relative to conjugate base in relation to the specific pH desired. One can achieve this specific ratio simply by adding a certain amount of strong base to adjust for the concentration of the anionic component. Note that such solutions will not change in relative concentration due to the common ion effect.

So use the henderson-hasselbach equation, solving for the concentration of the conjugate base that will be required, a fair amount of algebra is required.

9.0 = 9.6 + log(x/.100)

I got .0251 moles/L, this will be the final concentration achieved by adding sodium hydroxide.

Determine what the concentration of the conjugate base would have been in .100 M glycine without added base.

Use Ka = [H+][A-]/[HA], or Ka times [HA] = x squared

solve for x, I got .000005012 M, this is the original concentration of the conjugate base.

The question now is, what amount, or what relative ratio of volumes will we need to mix to go from the original to the final value?

concentration of the original V1 + concentration of sodium hydroxided added to the original V2 = the final concentration times (V1 + V2), solve for the ratio of volumes

I got 1/38.848, sodium hydroxide/glycine, meaning that we will mix .0051482702 L of sodium hydroxide with 1.948517298 L of glycine.

Ka = [H+][A-]/[HA] and substituting for this state equation, use the pH of 9.0 to find [H+], you'll see that the answer is correct.

7. Sep 18, 2004

### chem_tr

Thanks for this great explanation, GeneralChemTutor. But how many digits are we supposed to write anyway? I have nothing negative to say for your solving the problem.

Regards, chem_tr

8. Sep 19, 2004

### GCT

For the sake of accuracy, I had written all of the digits inclusive of the answer obtained on my calc., since the final answer involves a rather small amount of sodium hydroxide, it is best to be completely accurate. Which does not take too much work on your calc, since you can proceed directly from the answer obtained.

9. Sep 19, 2004

### RedVelvetCake

to chemtutor

hi chemtutor
i havent done chemistry in almost 8years, so what may seem simple to most does not seem easy to me:
Ive reviewed your solutions and Im not getting the same answers even with the explanation:::
mainly the first part for question 1 (ONLY)
9.0=9.6+ log(x/.100)
I dont get .0251moles/ L
Im getting .02826

this is how i solved please tell me what Im doing wrong:
(.100)9.0= 9.6 + log (x/.100) .100
.9 = 9.6 + log x
.9/9.6 = 9.6+logx/9.6
.0937= logx
.02826 = x

you and the book get .0251 but Ive tried several times and I get this
can you tell me what Im doing wrong?
thanks Nikki in NY

10. Sep 20, 2004

### chem_tr

Hello, I wanted to contribute to the problem a bit.

If the equation, $$9,0=9,6+log(\frac{x}{0,100})$$, is rearranged to give $$x$$, you'll need to use antilogarithm, after subtracting 9,6 from 9,0:

$$-0,6=log(\frac{x}{0,100})\Rightarrow10^{-0,6}=\frac{x}{0,100}\Rightarrowx$$, and then $$x=10^{0,6-1}=10^{-1,6}=0,0251$$

Regards, chem_tr

11. Sep 21, 2004

### GCT

Make sure to include all of the digits btw steps, don't estimate...this is probably where the problem lies.

12. Sep 23, 2004

### RedVelvetCake

.... hi chem tutor, im stuck again

with problem two Ive worked through it but im stuck at this point :

Ch3Cooh + H20 = CH3COO- + H30 +
1.0-x 0.5 +x x

Ka = ( 0.5+x ) ( x )
____________
1.0 - x

1.8 x 10-5 = (0.5 + x ) ( x)
___________
1.0 - x

13. Sep 23, 2004

### chem_tr

You've come a long way, but stuck in a point very close to the goal. Just omit the x in $$(1,0-x)$$, since there is a way too large difference between 1,0 M and 1,8*10-5 M. This will ease your calculations, since the denominator is terminated. You'll solve a quadratic equation now, by using the known formulae of discriminant $$\Delta=b^2-4ac$$, and so on.

14. Sep 23, 2004

### RedVelvetCake

thanks chem_tr.............

thanks chem_tr and chemtutor for being such a constant help to me, I dont mean to keep coming back but as I wrote before ,its been 8 years.

Thanks again I'll probably be back again for help
Nikki in New York

Last edited: Sep 23, 2004
15. Sep 27, 2004

### RedVelvetCake

please take a look at this

A 500ml sample of a .100M formate buffer, ph 3.75, is treated with 5ml of 1.00N KOH . What is the ph following this addition ?
IVe started but I dont know how to proceed and finish this :
HCOOH .1M
HCOO- .05 +x
H+ x

in the flask since ph = pka at 3.75 the concentration of A- and HA are equal at .05 M .
Ka= 1.1x10-3.75
my teacher got a value of .0495/505ml which i dont understand .and some value of .025-.0005/505 I dont understand how he got these values and I need to understand the steps to reach this conclusion : can someone please explain?
Nikki in NY

Last edited: Sep 27, 2004
16. Sep 27, 2004

### chem_tr

Hello

Formate buffer contains formic acid and formate in equimolar amounts. 250 mL of it is formic acid, whereas the rest belongs to formate. As you treat this buffer with 5 mmoles of potassium hydroxide, then some acid will convert to potassium formate.

$$HCOOH (25 mmol) \rightarrow HCOO^{-}(25 mmol)+H^{+}(25 mmol)$$
Don't forget to use the ionization constant. It contains 25 mmol of $$HCOO^{-}$$ more.
When you treat this buffer with $$OH^-$$, some more $$HCOO^-$$ will also be formed. If the base is sufficiently enough, then the pH will increase accordingly.

I will not give any more details, please study the problem a bit, and consult your books, after that, us, if you are stuck.