HELP! question about ph and pka!!!?

Question

So I have a ph of 2 and a molar concentration of .01 M of acid. how can i find the pKa?

what i wanted to do was:

pka = ph + log(concentration?)

pka = 2 + Log (2) = 0 .... but i can't remember much about this stuff and the formulas i have [such as ph = pka +log(A-/HA)] do not seem to work because i don't have a "base" concentration... don't know if i really make any sense at all.

well, i am at loss, so if someone can, HELP!!!! and quick!! Thanks!!

Answer 1

haha no problem finally im smart enough to use my AP chem knowledge.

pKa= -log Ka

umm appparantly im not as smart as i thought i was.  You cant find the pKa without one of these two things so see if you have them:

1) Acid(which u don't have)

2) Equation of the formula

if you have the equation, find Ka by doing the...

Wait, new idea. You have a pH so you have a pOH (base)

pOH=12 (14-2)  So now use the formula pH= -log (H) so do

10 ^ -pOH to get the molar concentration of the base (M)  Now do .01M/ (molar conc. of Base (M)) and use that formula

ph = pka +log(A-/HA).

pka=pH - log (ACID M/BASE M)

i hope that works

Answer 2

HA --> H+ + A-

pH = 2

[H+]=antilog -2=0.01M

[A-]=[H+]

Ka=[H+][A-]/[HA]=(0.01)(0.01)/(0.01)=0...

pKa=-logKa=-log0.01=2

in this case, pH=pKa