Calculating the Ka from pH and initial concentration

1. Nov 3, 2012

maceng7

1. The problem statement, all variables and given/known data
A 0.0200 mol/L acid solution has a pH of 2.347. Calculate the percent ionization of this acid. Calculate the Ka of this acid

3. The attempt at a solution

I first calculated the hydronium ion concentration: 10^-pH = 10^-2.347 = 4.5*10^-3

I then used this concentration to calculate the percent ionization: 0.0200 / 4.5*10^-3 = 22.5%

Can anyone confirm this answer. Now the second part of the question asks for the Ka. My prof said that if you got 1.3*10^-3 as the Ka value it is wrong. I keep getting this value and it makes no sense why it would be anything other than this value:

Since the ratio in the balanced chemical equation is 1:1 , the equilibrium expression is:
Ka = (4.5*10^-3)(4.5*10^-3) / (0.0200 - 4.5*10^-3)

solving for x I get 1.3*10^-3, however this isn't suppose to be the right answer.. can anyone tell me where I've gone wrong or maybe my prof is mistaken? Thanks.

2. Nov 3, 2012

symbolipoint

Your Ka value is wrong simply because you miscomputed your numeric expression. The numeric expression seems good; Try recomputing and you will, or should find a much smaller Ka value.

3. Nov 4, 2012

maceng7

I don't know I've put it into my calculator numerous times and I get the same value,

4. Nov 4, 2012

symbolipoint

You are incorrectly handling this computation on the calculator. Your expression seems to be written well and appears credible, so as a check, try doing the computation the old fashioned, manual way. You SHOULD find something about ?.?? x 10-6

5. Nov 4, 2012

Staff: Mentor

OK

No.

$$\frac {0.0200} {4.5\times 10^{-3}} = 4.44$$

Expression looks OK to me, but first - there is no x to solve for, second - as symbolipoint stated, it evaluates to something completely different.