# Homework Help: 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.