[bakerboy]

Hi,

While the question does involve amino acids, I am trying to solve it using gen. chem principals (so far to no avail), so I figured this would be the appropriate space.

The question is.

Estimate the pH of a solution of 300 mg of glycine (MW = 75, pI = 6.0 pKa1=2.3, pKa2=9.7) in 20 mL of water when you add 1.0mL of 1.0M H2SO4.

I know that I should be using Henderson Hasselbach(I think), but I do not know where/how to find the concentrations of the acid and base.

I am guessing the acid and base will be the protonated and neutral glycine--or would it be neutral and unprotonated glycine--but how do I find this?

Any suggestions would be much appreciated.

Thanks

bb

[Yggdrasil]

The amino acid glycine has three major forms.  From low pH to high pH:

H₃N⁺-CH₂-COOH
H₃N⁺-CH₂-COO^-
H₂N-CH₂-COO^-

The pKa values represent the transition between these forms.

[bakerboy]

Would this seem appropriate then?

MW of glycine = 75 g/mol
300 mg of glycine used.        Moles of glycine = .3g  x  1mol/75g = .004 moles of glycine
1.0 mL of 1.0M sulfuric acid moles of sulfuric acid =  1mol/L x .001L = .001 moles

Glycine- + sulfuric acid ⇌  glycine 0 + deprotonate sulfuric acid
.004      .001      0         0
-.001      -.001      +.001         +.001
.003      000      .001         .001

.001 moles of netural glycine (acid) /.021 L = .048M
.003 moles of negative glycine(base)/0.21L = .143M
When glycine interacted with water, it became negative, as it donated a proton to water (glycine inflection point lower than pH of water, therefore equilibrium shifts to compensate and create more negative glycines). With the sulfuric acid added, glycine became protonated and neutral again. This corresponds to the reaction occurring at pKa value of 9.7.
pH = pKa +  log ( [base]/[Acid] ) = 9.7 + log (.143/.0480) = 10.17

[Borek]

Completely off. Think this way - you mix two acids (one strong) and you get pH around 10 (basic solution). Could this be OK?

Start calculating pH of pure glycine and pure sufuric acid of given concentrations.

[bakerboy]

I thought the high pH was strange too.

Alright, attempt 2.

Finding the pH of glycine in 20mL of water, by using its pka of 2.3 to find its Ka, and from its Ka finding its H+ concentration using an Ice Table
HA<-> H+ A-
.004   0   0
-x      +x  +x

Find x to be 0.0026. and find the pH to be 2.58.

Then, pH of 1M sulfuric acid is -log (1) = 0.

From this, we can find hte [H+] of both, and because we know the volume added, we can determine the amount moles of H+. Add the moles together, divide by sumated volume, and find new pH of acid.

10^-2.58 = .0026 M H+ from glycine x.02 L of water = .000052 moles of H

1M of H from sulfuric acid x .001 L = .001 moles of H

.001moles + .00052 = .001052 moles of H /0.021 L = .0500M H+

pH = - log (.0500) = 1.30


While I could believe this, you get a slightly different answer when you use the  pKa2 for glycine--the H concentration is much smaller--but due to the large amount of moles from sulfuric acid, it doesn't make too much difference in the end pH roughly 1.32.

Thanks for the help

bb

[Borek]

H⁺ concentrations isn't simply additive, but as long as there is reasonable excess of sulfuric acid, glycine presence doesn't change situation - much.

Please remember, that sulfuric acid is diprotic with pK_a2 = 2.

[eheheh]

I'm trying to solve the same problem ... i did this :

0.300g of glycine X 1 mol/ 75 g glycine = 0.004 mol glycine

0.001L H2 SO4 X 2 mol/L H2 SO4 = 0.002 mol H+ added to the solution

Given  that all H+ will react with glycine to form HA, then HA will be the same number of mols as H+ added to the solution. Therefore, H+ = HA = 0.002

 (0.004 mol glycine solution – 0.002mol H+) / (0.0200L + 0.001L) = 0.0952 M A-


pH    = pK1 + log [A-]/ [HA]

   = 2.3 + log 0.0952M/ 0.002M
   = 2.3+ 1.678
            = 3.98
But this answer makes no sense.. it should be around 1.4-2..
what did I do wrong?!

[Borek]

Arrgh, seems you can be right... I think I have miscalculated glycyne concentration.

[bakerboy]

In response to eheheh:

You forgot convert HA to Molarity, you left it in moles. It should be .002 moles/.021L , which will equal the same thing as [A-] making hte pH the same as the pKa.

I am definitely not hte person to say whether or not the approach is right though. haha

Borek:

Am I on the right track with the wrong method?

[Borek]

You forgot convert HA to Molarity, you left it in moles. It should be .002 moles/.021L , which will equal the same thing as [A-] making hte pH the same as the pKa.

And that's the correct answer. I think. I was already wrong twice, I can be wrong for the third time.

Well... it is not exactly right. HSO₄^- is not strong enough, so assumption that there is 2 mmol of H⁺ from sulfuric acid dissociation is wrong. But at least approach is correct this time. 2 mmoles of acid protonating 4 mmoles of glycine, giving 1:1 ratio of protonated and neutral molecule, giving pH=pKa.

I should punish myself for being wrong. Trick is, I have no idea what should the punishment be. I can either calculate exact pH value, or not calculate it. In the first case, I will punish myself with additional work. In the second case, I will punish myself cause I like to analyse these things 

[Borek]

Did the exact calculation with Buffer Maker.