Annoying acid/base problem...

[unitix]

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Okay, I'm a gen chem buff and usually destroy PS on practice tests. But like most others I have some acid/base/buffers anxiety but am still relatively good at them. This one passage-less problem on a past kaptest really bugs me as I can't seem to recall what I am missing here...

When 10mL of 1M K3PO4 solution is titrated with 1M HCl, the pH corresponding to the hydrophosphate/dihydrophosphate buffer region is found to be "X". What is the dissociation constant of dihydrophosphate?

A. 10^(1-X)
B. -log10^-X
C. 10^-X
D. 10^(1+X)

How I went about it...

pH = X = pKa + log(b/a)....pKa ---> -logKa
then raise everything onto 10 cause of the logs...

10^X = 10^(-logKa) + 10^log(b/a)

log rules from like 7th grade...(10^logA = A)

10^X = 1/Ka + (b/a)

easy parts over, now with the confusion...the answer is C. Ka = 10^-X

I see that somewhere, somehow "b/a" had to have been assumed 0..?

10^X = 1/Ka + 0 -----> Ka = 10^-X

wtf am I missing here

 

[MedPR]

The buffering region (where pH deviates the least, hence "buffer") is where pH=pKa. If pH=pKa, then logb/a must be 0, in other words, moles conjugate base = moles acid. You might also know this region as the "half-equivalence point."

This picture is backwards, since it is weak acid and strong base, but it is the same idea as your problem.

 

[chiddler]

"A. Ka = 10^-X"

is the answer A or C?

 

[MedPR]

"A. Ka = 10^-X"

is the answer A or C?

Should be C, right?

 

[chiddler]

Should be C, right?

i hope. it's what i got.

 

[MedPR]

i hope. it's what i got.

Cool, me too 🙂

X=pKa + 0

X=-logKa + 0

10^X=10^(-logKa)

10^X=1/Ka

Ka*10^X=1

Ka=1/10^X

Ka=10^(-X)

 

[unitix]

EDITED: answer IS C lol SORRY guys haha

and OMG I knew it was a little thing I was forgetting....pH=pKa in buffer region DUH

Thanks guys