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jkn1121

Biochem Question

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  • The aminoacid glycine is often used as the main ingredient of a buffer in biochemical experiments. The amino group of glycine, which has a pKa of 9.6, can exist either in the protonated form ( -NH3+) or as the free base (-NH2), because of the reversible equilibrium

R-NH3+ Û R-NH2 + H+

a) In what pH range can glycine be used as an effective buffer due to its amino group

 

B) In a 0.1 M solution of glycine at pH 9.0, what percentage of glycine has it’s amino group in the NH3+ form?

 

c) How much of 5 M NaOH must be added to 1 L of 0.1 M glycine at pH 9.0 to bring its pH to exactly 10.0?

 

d) When 99% of the glycine is in its –NH3+ form, what is the numerical relation between the pH of the solution and the pKa of the amino group? (for example, “the pH is double of the pKa” or “the pH is 1 pH unit above the pKa”, or “pH = pKa + 1”, etc.)

 

I dont know where to begin.

 

 

Edited by jkn1121

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You might start by reviewing the concept of acid dissociation, the dissociation constant and the relationship to pH. Perhaps you might start by indicating what you know of this. You might also start by suggesting which of the four questions you understand best and how you might approach that particular problem.

 

Have you ever worked chemical equilibrium problems?

Edited by cypress

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yea, i know that pka 9.6 is like the pH is 2.511x 10^-10 which quite small and the pka is basic so the the pOH is 4.4 which is 3.98x 10^-5.

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OK let's answer the first question.

 

When you rearrange the formulas to derive the relationship between pH and pKa you get this one:

 

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

 

So at a pH equal to pKa what can you say about the ratio between [A-] and [HA] ? with respect to the amino group on glycine, what is the equivalent of A-? and HA? What can you say about the ability of glycine to buffer OH (donate H+) beyond this point?

 

Once you have this figured let's go the other way and you tell me where we should expect the amino group to stop accepting H+.

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You have to find the concentration of A, HA, in order to solve fo pH if the equation is pH = pKa + log ([A-]/[HA]). So for Part A the solution is quite basic. It can range from 7-14. Then using the 0.1M in pH= 9.0 thats consider the A in the formula, so you plug it in, to get 9= 9.6 + log (0.1/HA)?

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You have to find the concentration of A, HA, in order to solve fo pH if the equation is pH = pKa + log ([A-]/[HA]). So for Part A the solution is quite basic. It can range from 7-14.

 

Actually it can range from 1-14, but don't lose sight of what the question is asking. It is asking about glycine as an "effective buffer", and buffers, well they buffer changes in pH by donating or accepting H+ as needed to make the solution less susceptible to pH changes as H+ or OH- is added to the solution. So look at the equation I offered again and predict where glycine's amino group loses it's ability to donate more H+ to counter the effects of adding OH- and thus raising the pH. Next do the same as H+ is added to reduce pH.

 

 

Then using the 0.1M in pH= 9.0 thats consider the A in the formula, so you plug it in, to get 9= 9.6 + log (0.1/HA)?

 

Close but remember [A-] + [HA] is 0.1 M and while A- represents the NH2 form of the amino group, it would be more like 9=9.6+log((0.1-[HA])/[HA]) right?

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Ok, The formula is pH = pKa + log ([A-]/[HA]) (I assume you understand how this formula was derived, but please ask if you do not) implies a ratio of [A-] to [HA] when pH = pKa because at that point log([A-]/[HA]) is zero. Notice what this ratio is and then notice what the ratio is if log([A-]/[HA]) is +1 and -1 respectively so that pH is 10.6 and 8.6. Note that as the amino group on glycine donates H+ it reverts to the NH2 form or A- form and as it accepts H+ it takes the NH3+ or HA form.

 

Based on this, how much further in either direction of the pH scale is glycine likely to donate or accept H+ significant amounts of H+?

 

How about question b ) did you get the percentage?

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So you are saying in A. ranges from pH is 10.6 and 8.6. As for B pH = pKa + log ([A-]/[HA]) 9= 2.51x 10^-10 + log (A/HA)?

 

 

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So you are saying in A. ranges from pH is 10.6 and 8.6.

 

AI'm trying to get you to select the range and explain the basis for the choice. 8.6 to 10.6 was a range I selected... What do you select and why?

 

As for B pH = pKa + log ([A-]/[HA]) 9= 2.51x 10^-10 + log (A/HA)?

 

Ka is 2.52x10^-10 but pKA is 9.6 so I get 9 = 9.6 + log ([g-NH2]/[g-NH3+]) so -0.6 = log((0.1-[g-NH3+])/[g-NH3+]) then

 

((0.1-[g-NH3+])/[g-NH3+]) = 10^-0.6 = .2511 so [g-NH3+] is what % of the total?

 

edited back when I realized that in my rush I forgot what I was doing. It was correct the first time.

Edited by cypress

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A. I think the pH range as of 9.6 from 9-10 for the buffer since the buffer keeps the system at equilibrium and the pH doesnt change that much when the buffer is added it.

B. If I piggyback on what you did, the concetration is .2511 [-NH3] then it should be 75% NH3, 25% NH2

 

 

 

I don't unerstand how it went from ([g-NH2]/[g-NH3+]) to ([g-NH3]/[g-NH3+])

 

 

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A. I think the pH range as of 9.6 from 9-10 for the buffer since the buffer keeps the system at equilibrium and the pH doesnt change that much when the buffer is added it.

 

Not a bad answer, but it is subjective and the pH range is unjustifiably unbalanced around the midpoint. I was trying to get you to see that at a pH of 9.6 the concentration of g-NH2 is equal to the concentration of g-NH3+ because when the pH = pKa then the log component is zero and since log(1) = 0 , the ratio of the components is 1 so each component is in equal concentration at that point. Notice that when the log component is log(10/1)=1 or log(1/10) =-1 then the proportion of corresponding components is 1 to 11 or in other words 91% of the buffering capacity is consumed for a change pH change of 1. At what pecentage would you consider the buffering capacity mostly consumed?

 

 

B. If I piggyback on what you did, the concetration is .2511 [-NH3] then it should be 75% NH3, 25% NH2

 

Ill come back to this. got to eat now.

 

I don't unerstand how it went from ([g-NH2]/[g-NH3+]) to ([g-NH3]/[g-NH3+])

 

Yes, thanks ... sorry for the typo... my mistake. .

 

...edit: no sorry I was confused. It is correct as is... the change is that it

went from ([g-NH2]/[g-NH3+]) to (0.1 - [g-NH3+])/[g-NH3+]) 0.1 is the concentration of all the glycine.

[g-NH2] + [g-NH3+] = 0.1 (all the glycine) so (0.1 - [g-NH3+]) = [g-NH2]

Edited by cypress

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Would Part C besince 5M*1L= 5moles

10= 9.6 log(5/HA) , so then

.4= log (5/HA) then

10^ -0.4 = 5/HA, to

2.51 (HA)=5 then

A =2?

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B. If I piggyback on what you did, the concetration is .2511 [-NH3] then it should be 75% NH3, 25% NH2

 

Careful with your math.... 0.25/0.75 is not = .2511. Notice that it is ~1/4 so it is about 0.02/0.08 or 80% g-NH3. this is why you must use (0.1-[g-NH3+])/[g-NH3+]

because the two together equals the whole. Your mistake is common. Make sure you don't make that one on a test.

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So are you saying that the pH range can be from 1-11 since the concentrations are equal to each other?

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So are you saying that the pH range can be from 1-11 since the concentrations are equal to each other?

 

No, when the pH is 1 unit higher or one unit lower than pKa, in other words when the pH is 8.6 or 10.6 then the ratio of the two components will be 1 part to 10 parts (1/10) or (10/1) and the ratio of the whole is 1 to 11 and one out of 11 is about 9%. Thus 9% of the buffering capacity remains (91% is consumed) when the pH is one unit away from pKa.

 

Notice realized I misread your previous question and thought I made an error. I restored it to the original and explained the change hope you are following it.

 

Would Part C besince 5M*1L= 5moles

10= 9.6 log(5/HA)

 

You're headed in the right direction but you made the same mistake as in part b and you are starting at the wrong spot. You don't start with the NaOH you start with the solution you have. Let's take a step back....

 

You begin with 0.1 molar solution of glycine at pH=9.0 so that is 0.080 moles of NH3+ (from the previous problem)

You end with 0.1 molar solution of glycine at pH=10.0 so that is what concentration of NH3+ ?

 

10.0 = 9.6 + log((0.1 - [g-NH3+])/[g-NH3+]) note the same formula switch as before

 

So do the math and let me know what end concentration you get for g-NH3+

 

The difference is the molar amount of OH you need to add.

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0.4= log ((0.1 -[g-NH3+])/[g-NH3+])

Anti Log

.398 = 0.1 [(g-NH3+])/[g-0.080+])

Edited by jkn1121

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I get 10^(0.4) = 2.512

 

edit: by the way g-NH2 refers to the glycine in the NH2 form and g-NH3+ is the glycine in the protonated form. It is not glycine minus NH2. Sorry if this was confusing.

Edited by cypress

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Ok, I see what I did... I added the negative sign, so once I get

2.512= 0.1 -[(g-NH2+])/[g-NH3+])

 

would you then divide?

 

Is the 0.080 part of theNH3+ conc.?

Edited by jkn1121

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[g-NH3+] means "concentration of glycine in the protonated form"

 

We are solving for the final concentration first then we will subtract the final from 0.080 which is the beginning right?

 

You need to rearrange the equation to solve for the concentration of g-NH3+

 

2.512*[g-NH3+] = 0.1 - [g-NH3+] ----> [g-NH3+] = 0.1/3.512

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I think I get up to the point where its :

2.512 = 0.1 -[(g-NH2+])/[g-NH3+])

then you cross multiply to get

 

2.512*[g-NH3+] =0.1- NH2,

 

 

I dont understand [g-NH3+] = 0.1/3.512 part

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I think I get up to the point where its :

2.512 = 0.1 -[(g-NH2+])/[g-NH3+])

then you cross multiply to get

 

2.512*[g-NH3+] =0.1- NH2,

 

 

I dont understand [g-NH3+] = 0.1/3.512 part

 

I see the confusion.... Remember the original solution is 0.1 molar glycine so [g-NH2] plus [g-NH3+] together always totals 0.1 unless more glycine is added.

 

so [g-NH2] = (0.1-[g-NH3+]) and the formula is not as you wrote it,

rather it is 2.512 = (0.1 - [g-NH3+])/[g-NH3+]

 

edit: to see this notice that the original equation is

 

2.512 = [g-NH2]/[g-NH3+] but [g-NH2] = (0.1-[g-NH3+]) so

 

2.512 = (0.1 - [g-NH3+])/[g-NH3+]

 

I put it in this form to solve it for [g-NH3+] .... you know, one equation one unknown....

Edited by cypress

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So you have 2 equations:

 

 

2.512 = (0.1 - [g-NH3+])/[g-NH3+]

 

 

[g-NH2] = (0.1-[g-NH3+]

 

________________________________

 

[g-NH2] = (0.1-[g-NH3+]

 

= (0.1- .080)

= 0.02

2.512= 0.1- ( 0.080/[g-NH3])

2.512/0.02

= 125.6

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So you have 2 equations:

 

 

2.512 = (0.1 - [g-NH3+])/[g-NH3+]

 

 

[g-NH2] = (0.1-[g-NH3+]

 

________________________________

 

 

Yes, good but don't mix the values 0.08 is the concentration of the NH3 form of glycine at a pH of 9.0 and we are solving for the concentration of the NH3 form at a pH of 10.0

 

 

[

g-NH2] = (0.1-[g-NH3+]

 

= (0.1- .080)

= 0.02

 

Yes fine this is the concentration of the NH2 form at pH of 9

 

2.512= 0.1- ( 0.080/[g-NH3])

2.512/0.02

= 125.6

 

No, here you mixed the 9.0 pH concentration into the formula for the 10.0 pH concentration.

 

2.512 = (0.1-[g-NH3+])/[g-NH3+] ---> 2.512*[g-NH3+] = 0.1 - [g-NH3+] ---->

 

2.512*[g-NH3+] + [g-NH3+] = 0.1 ---> (2.512 + 1)*[g-NH3+] = 0.1 ---> [g-NH3+] = 0.1/3.512

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so the final answer is 0.028 of 5M NaOH needed to be added

 

 

Edited by jkn1121

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