titrating amino acid

[cloak25]

How many mL of 0.010 M NaOH need to be added to take 10 mL 0.010 M lysine from pH = 5.68 (the first equivalent point) to pH = 10.80 (pKa3 for lysine)?

A. 5 mL 0.10M NaOH
B. 10 mL 0.10M NaOH
C. 15 mL 0.10 M NaOH
D. 20 mL 0.10M NaOH

Answer: C

BR solved this using a titration curve graph but can you do this intuitively? mathematically?

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[pfaction]

I got this wrong. I know equivalence is moles = moles, so I tried to do it that way. I ended up getting B. Also, is this 0.010 or 0.10?

 

[chiddler]

pH = pKa 3 is when conj. Base = Acid. Since we start from equivalence #1, we add one equivalent of base to reach equivalence point 2. And then a little bit more to reach equivalence point 2.5 (the half equivalence point) when base = acid. So answer is more than one equivalent, but less than two equivalents. C fulfills this.

 

[pfaction]

What's this about "half equivalence point"??

 

[chiddler]

What's this about "half equivalence point"??

lol like i wrote it's when pH = pKa

 

[pfaction]

Yeah I get that. When you add half volEquiv you'll get pH = pKa. I understand I made that mistake. But the 2nd pH = pKa point is less than the 2nd equivalent, isn't it? It's half an eq less. So I don't get what you mean at all.

You start at 5.7, adding an equivalent will bring you to the 2nd equivalent, which is higher than 10.8, you want half an eq lower, right?

EDIT: HOORAY READING COMPREHENSION;

SAYS PKA3 LOOOOOOOL



So let me do the math.

To go from the first equivalent to the 2nd equivalent will be, and I'm assuming you are using 0.1M NaOH:

Lysine = 0.1/1000 = x/10; x=0.001 moles.
NaOH is the same thing; therefore 0.001 moles = 10mL to get to the 2nd equivalent.

Since pH = pKa, and VEq/2 will give you this, we have to add half an equivalent: 10+(10/2) = 15.

Thank you chiddler, and you can now see why the **** I'm going to fail this MCAT, hard.

 

[chiddler]

Thank you chiddler, and you can now see why the **** I'm going to fail this MCAT, hard.

=/

 

[pfaction]

Seriously, how ****ed up am I when I read pKa for equivalence, and then misread 3 for 2?

Also, an alternate way thanks to chiddler's shortcut which I just realized:

Since you'll need 20mL = 2 equivalents (from 1 to 3), and half equivalent is a pH = pKa, the 3rd pKa is inbetween the 2nd eq and 3rd eq, so it's gonna be less than 20, but more than 10....good jorb.

 

[chiddler]

Seriously, how ****ed up am I when I read pKa for equivalence, and then misread 3 for 2?

i think we all do those mistakes occasionally