Biochemistry experts, help

[tdkyun]

Hello,
I am currently studying for the DAT and taking biochemistry. There was a quiz in my biochem class and I missed this problem and I need your help.

One liter of an aqueous solution contains 0.050 M asparagine at its isoelectric point. Given that pK1=2.0 and pK2=8.8, how many moles of strong base must be added (assume constant volume) to raise the pH to 9.0?
A. 0.010
B. 0.020
C. 0.030
D. 0.040
E. 0.050

Thanks in advance.

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[tdkyun]

bump, anyone? =[

 

[mathcod]

I'm probably not going to solve it, but I'll try to think it out with you.

pH(initial) = (2.0 + 8.8)/2

@ pH(initial), 100% in acid form between (0/-)

Molarity * 1L = #mols

#mols acid form(initial) = #mols asp

want pH(final) = 9

9 = pKa (which is 8.8) + log (b/a)
solve for ratio of molecules in b and a forms (b form is 61% of the molecules)

so in total, b constitutes .61*.05mol = .030656mol in base form


assuming 1M strong base, add the volume of base that would yield enough mols of base to move from the initial to the final state, which would be .030656mols of strong base should do it.

I think that's the right process. if you've any idea about problems like this, you should be able to understand what Ive done. biochemistry is a great class. I loved it.

It was a huge hurdle for me to understand biochem, but once it clicks, you'll remember it forever! If you have trouble, see a ta or tutor. This is fundamental acid/base chemistry so it's worth learning. (plus its cool)

If you don't follow what I've done, or I did it incorrectly, others can help you out. cheers

Actually i did end up solving it cuz it was a good mental exercise. Thanks for the fun.

 

[tdkyun]

do you mind explaining how you got 61%?
thank you.

 

[mathcod]

Henderson-Hasselbach

pH = Pka + log b/a

b and a can be either in molarity or mols, it doesn't matter but they both have to be in the same units because the log of a number must be unitless!

9 = 8.8 +log b/a

.2 = log b/a

10^.2 = b/a

10^.2
_____ =
1

b
_
a


what i just did was expressed the number divided by 1 (which is of course the same number) to show the ratio that for every 10^.2 molecules of base form, there is 1 molecule of acid form

now to get the % of molecules in base form, do (B/total) which is [B/(B+A)], which in this problem is (10^.2)/(10^.2 + 1) =.61

this means that 61% of the total species is in base form

to get the % of molecules in acid form, do (A/total) which is A/(B+A),
or you can just do 1 - % molecules in base form or (1-.61) because both species have to represent 100%.

We only calculated the % of the species in acid and base forms, so to get the number of moles in either form, multiply the %'s by the moles of species.