[rproctor]

A 0.5 M solution of benzoic acid has a pKa of 4.19

a) what is the pH of this soultion

b)What is the % of benzoic acid ionized in this solution?

[Jeremy]

You need to show your attempt before anyone can help you.

[rproctor]

Sorry!

First I converted the pKa to Ka. I found Ka to be 6.46 x 10^-5. I then created an ICE table and solved for x finding it to be .00568. I  found that pH= 2.25

Did i do that correctly?

As far as part B, i am not really sure how to go about answering that.

[UG]

The first part looks good.
For the second part, the % acid ionisation is simply the amount of conjugate base formed divided by the original amount of benzoic acid. You have all this information when working out the first part.

[rproctor]

I found that the % of benzoic acid ionized to be 11.4%. Does that sound accurate?

[UG]

I found that the % of benzoic acid ionized to be 11.4%. Does that sound accurate?

From the first part, the amount of conjugate base formed should be equal to the amount of H⁺ so therefore = 10^-2.25 = 5.6234x10^-3 M.
Then % ionisation should be 5.6234x10^-3/0.5 x 100%