Acid/base chemistry question

[escesc]

I would appreciate someones help with this:

The approximate pKa values for the three dissociating groups of lysine are 2.2, 9.0, and 10.5. What would be the pH after addition of 10ml of 1.0M NaOH to 100 ml of .1 M lysine at pH 2.2?

Thank you.

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[BerkReviewTeach]

I would appreciate someones help with this:

The approximate pKa values for the three dissociating groups of lysine are 2.2, 9.0, and 10.5. What would be the pH after addition of 10ml of 1.0M NaOH to 100 ml of .1 M lysine at pH 2.2?

Thank you.

If it's 10 mL of 0.1 M NaOH and 100 mL of 0.1 M lysine, then you have equal mole quantities of the two species (10 x 1.0 = 100 x 0.1). Starting at 2.2, the pH it will go up by one equivalent on the titration curve. At pH = 2.2, it's half way between the start and first eq point. After one full equivalent, it will go up to half way between the first eq point and second eq point, which is where pH = pKa2. The pH would be about 9.0.

The curve would have the following key points:

• 
  Start at pH = 1.1
  1/2 equiv: pH = pKa1 = 2.0
  First equiv: pH = (pKa1 + pKa2)/2 = (2.0 + 9.0)/2 = 11/2 = 5.5
  1 1/2 equiv: pH = pKa2 = 9.0
  Second equiv: pH = (pKa2 + pKa3)/2 = (9.0 + 10.5)/2 = 19.5/2 = 9.75
  2 1/2 equiv: pH = pKa3 = 10.5
  Third equiv: pH = pKa3 + pHbase titrant = (10.5 + 14.0) = 24.5/2 = 12.25 (this is a slight over-approximation)

 

[escesc]

Thanks a lot for your help. For the question below will it also shift one pka making it 8.9?

The pka values for three dissociating groups of Lysine are 2.1, 8.9, and 10.6. What would be the pH after addition of 10ml of 2.0 M HCL to 10 ml of 2.0 M Lysine at pH 10.6?

 

[BerkReviewTeach]

Thanks a lot for your help. For the question below will it also shift one pka making it 8.9?

The pka values for three dissociating groups of Lysine are 2.1, 8.9, and 10.6. What would be the pH after addition of 10ml of 2.0 M HCL to 10 ml of 2.0 M Lysine at pH 10.6?

Exactly!

 

[premed101]

If it's 10 mL of 0.1 M NaOH and 100 mL of 0.1 M lysine, then you have equal mole quantities of the two species (10 x 1.0 = 100 x 0.1). Starting at 2.2, the pH it will go up by one equivalent on the titration curve. At pH = 2.2, it's half way between the start and first eq point. After one full equivalent, it will go up to half way between the first eq point and second eq point, which is where pH = pKa2. The pH would be about 9.0.

The curve would have the following key points:

• Start at pH = 1.1
  1/2 equiv: pH = pKa1 = 2.0
  First equiv: pH = (pKa1 + pKa2)/2 = (2.0 + 9.0)/2 = 11/2 = 5.5
  1 1/2 equiv: pH = pKa2 = 9.0
  Second equiv: pH = (pKa2 + pKa3)/2 = (9.0 + 10.5)/2 = 19.5/2 = 9.75
  2 1/2 equiv: pH = pKa3 = 10.5
  Third equiv: pH = pKa3 + pHbase titrant = (10.5 + 14.0) = 24.5/2 = 12.25 (this is a slight over-approximation)

Although a moot point, it may be useful to clarify the bold is "1M."
Also, in your curve, why did you start with pH=1.1 if the pH of the solution is 2.2?
Lastly, how was the 3rd equivalent calculated?(specifically the pH base titrant)

 

[escesc]

Exactly!

I got the test key and it says it is 2.1 rather than 8.9. Is it incorrect?

 

[premed101]

Are you sure you copied down the question exactly?

 

[BerkReviewTeach]

Although a moot point, it may be useful to clarify the bold is "1M."
Also, in your curve, why did you start with pH=1.1 if the pH of the solution is 2.2?
Lastly, how was the 3rd equivalent calculated?(specifically the pH base titrant)

(1) Yeah, you are right that I mis-copied the molarity from the original post. Luckily the math used the right value, because that's much more than a moot point on that typo.

(2) The numbers I put were for the titration starting from Lysine in its fully protonated state, not the starting point listed in the question. The question correlates to the first pKa point on the titration curve, not the start of a full titration curve.

(3) The pH at the third eq point is a TBR shortcut. The last equiv point can be approximated quickly as (pKa3 + pH of the titrant)/2 - 0.15. The averaging is because of the symmetry of the curve (just like we do when determining isoelectric points) and the subtracting of 0.15 is an approximation due to the dilution factor during titration. And using pH = 14 for the titrant is because the titrant was 1.0 M NaOH.

 

[FlipDoc2Be]

If it's 10 mL of 0.1 M NaOH and 100 mL of 0.1 M lysine, then you have equal mole quantities of the two species (10 x 1.0 = 100 x 0.1). Starting at 2.2, the pH it will go up by one equivalent on the titration curve. At pH = 2.2, it's half way between the start and first eq point. After one full equivalent, it will go up to half way between the first eq point and second eq point, which is where pH = pKa2. The pH would be about 9.0.

The curve would have the following key points:

• Start at pH = 1.1
  1/2 equiv: pH = pKa1 = 2.0
  First equiv: pH = (pKa1 + pKa2)/2 = (2.0 + 9.0)/2 = 11/2 = 5.5
  1 1/2 equiv: pH = pKa2 = 9.0
  Second equiv: pH = (pKa2 + pKa3)/2 = (9.0 + 10.5)/2 = 19.5/2 = 9.75
  2 1/2 equiv: pH = pKa3 = 10.5
  Third equiv: pH = pKa3 + pHbase titrant = (10.5 + 14.0) = 24.5/2 = 12.25 (this is a slight over-approximation)

I'm confused. Could someone please explain why the pH would be 9?