Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Sis290025 on August 24, 2010, 05:44:03 PM
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Glycine is often used as a buffer to assay the enzyme pepsin. It has two pKa values: 2.34 and 9.6. In an experiment using a buffer containing 0.05 M glycine at pH 1.7, the optimum pH for this assay, it turned out that 5 mL of 5 N NaOH fell into 1 L of the buffer. What’s the resultant pH?
Ionization of glycine results in three forms, which I will identify as:
A - both the amino and carboxyl groups are protonated
B - the carboxyl group is deprotonated
C - both groups are deprotonated
Given that the glycine concentration is 0.05 M, is it correct to state:
[A] + {B} + [C] = 0.05 M ?
Therefore, if the above is true, then can it be established from the Henderson-Hasselbalch equation:
{B}/[A] = antilog (1.7 – 2.34) and
[C]/{B} = antilog (1.7 – 9.6) ?
Thank you.
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You are on the right track.
pH is so low, that means you may assume form C concentration is very close to zero, which in turn means
[A] + [ B] = 0.05
This is only an approximation, but it will work perfectly.
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The following depicts solving for [A] and {B}.
If .22909 = {B}/[A] and 1.259*10^-8 = [C]/{B}, then [A] = {B}/(0.22909) and [C] = (1.259*10^-8)*{B}.
Thus,
{B} + {B}/.22909 + (1.259*10^-8)*{B} = 0.05 M.
Solving for {B} gives 0.009319 M, which is used to find [A] = 0.04068 M.
After NaOH (0.025 mol) is added to 1 L, the new setup is:
pH = 2.34 + log [(0.009319 mol + 0.025 mol)/(0.04068 mol – 0.025 mol)] = 2.68 ???
Are my steps for the pH correct? If not, please point me in the right direction.
Thank you.
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I haven't checked the numbers, but approach seems OK.
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A chemist has a 50 mL solution of 25 mM histidine at pH 0. How many equivalents of NaOH does it take to titrate histidine to pH 12.5?
For the above question, must I know histidine’s pKa values as there are four forms of histidine regarding ionization?
Would I then follow similar steps as in the glycine example to calculate each form’s concentration at pH 0 (where [A] + {B} + [C] + [D] = 25 mM) so that I may use the Henderson-Hasselbalch equation for the appropriate species pair at pH 12.5?
Does the term “equivalents” refer to the volume of NaOH? If not, then what is mean by “equivalents”?
Thank you.
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You don't need all pKa values, just the first (to calculate initial concentrations) and the last (to calculate the final). Other than that question is not different from the previous one.
You may use [ B] - put space after "[ ".
http://en.wikipedia.org/wiki/Equivalent_(chemistry)
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Could you please clarify your first statement?
Histidine’s pKa values are 1.82, 6.00, and 9.17. Are you saying that I only need the values 1.82 and 9.17? Could you please explain or define by what you mean by “initial concentrations” and “final concentrations”?
Once again, thank you for your time.
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Histidine’s pKa values are 1.82, 6.00, and 9.17. Are you saying that I only need the values 1.82 and 9.17?
Yes, you won't use 6.00 in your calculations. Well, if it was very close to 9.17 if would be important, but it is not.
Could you please explain or define by what you mean by “initial concentrations” and “final concentrations”?
Initial - A, B at pH 0, final - C, D at pH 12.5.
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[ B] / [A] = 10^(-1.82) initial
[D]/[C] = 10^(3.33) final
But how am I to solve for the individual concentrations A, B, C, and D if I do not have a common species as I did in the equations/ratios for glycine? Is it safe to assume that at pH of 0, species A and B are predominant and, consequently, [A] + [ B] = 25 mM? Or, at pH 12.5, the predominant species are [C] and [D] and, thus, [C] + [D] = 25 mM?
Once these concentrations are calculated, how may I apply the information to finding the amount of NaOH added?
Originally, I thought that once I found the A, B, C, and D concentrations at pH 0 I would then be able to set up the equation as:
12.5 – 9.17 = log[(mol of D + x)/(mol of C – x)],
where x = mol NaOH added.
Thank you.
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Is it safe to assume that at pH of 0, species A and B are predominant and, consequently, [A] + [ B] = 25 mM? Or, at pH 12.5, the predominant species are [C] and [D] and, thus, [C] + [D] = 25 mM
Yes.
See equation 2 at http://www.titrations.info/acid-base-titration-indicators - while it is aimed at indicators, it holds for any weak acid/base. What it tells you is that when you are 2 pH units from pKa concentration of one forms is 100 times higher than the concentration of the other.
Once these concentrations are calculated, how may I apply the information to finding the amount of NaOH added?
Write reaction equations - you are very close in your approach, you just have to take into account 1 additional neutralization step. Think in terms of simple stoichiometry how much NaOH you will need.
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I have solved for A, B, C, and D in terms of moles after finding the total number of moles of histidine.
A = 0.001231 mol
B = 0.000019 mol
C = 5.84*10^-7 mol
D = 0.001249 mol
For the neutralization reaction regarding carboxyl group, see below:
RCOOH + NaOH --> RCOO- + H2O
0.001231 mol RCOOH will react completely with base in which base is represented by x – 0.001231 mol, where x is original mol of NaOH.
Thus, can I then use the following setup in which I solve for x?
12.5 – 9.17 = log[(mol of D + x – 0.001231 mol)/(mol of C – x + 0.001231 mol)]
Would mol of NaOH be equal to equivalents of NaOH?
Thank you.
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D = 0.001249 mol
Must be wrong. Should be smaller than C.
Would mol of NaOH be equal to equivalents of NaOH?
Yes.
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Well, since the pH (12.5) > pKa (9.17), wouldn't the predominant species be the depronated amino group form, or D?
Thank you.
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You have listed four numbers, with A >> B >> C - that suggests you did your calculations for low pH. At this low pH D is not predominant species. Then A ≈ D - there is no pH at which both A & D can dominate, so no matter how you look at these numbers, they are inconsistent.
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For A and B, the calculations were done at pH 0, but, for C and D, calculations were done at pH 12.5 for final.
Note the addition of A and B (and, hence, the sum of D and C) were set to to moles of solution instead of molarity.
If 10^3.33 = [D]/[C] is correct at pH 12.5, then [D] = 2138[C].
Could you please tell me where my calculations start to veer off?
Thank you.
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For A and B, the calculations were done at pH 0, but, for C and D, calculations were done at pH 12.5 for final.
If so, perhaps these numbers are correct (at least they don't look illogical). But you have to be a bit more specific in future, as this
I have solved for A, B, C, and D in terms of moles after finding the total number of moles of histidine.
A = 0.001231 mol
B = 0.000019 mol
C = 5.84*10^-7 mol
D = 0.001249 mol
is not clear.