[dantex3]

I hv finished other questions



   A solution is prepared by dissolving sodium benzoate and benzoic acid in water to give a 9 mmol/litre concentration of each compound. The mixture has a pH of 4.21 at 25oC. Calculate the acid dissociation constant of benzoic acid.
   Hint:   pH = pKa + log10 ·[sodium benzoate] / [benzoic acid]

[JGK]

Let me hazard a guess, you haven't read the FORUM RULES Have you? 

[AWK]

Start learning from Borek primer
http://www.chembuddy.com/?left=pH-calculation&right=toc

[dantex3]

Let me hazard a guess, you haven't read the FORUM RULES Have you? 

sorry , so i have try to do...and finish it...
but having 1 question can't solve

[Olshia]

[H⁺]= 10^-pH

[H⁺]=ROOT(K_a[ACID])

K_a=[H⁺]²/[ACID]

If you combine those equations you should get your answer.

[Borek]

[H⁺]=ROOT(K_a[ACID])

Approximation only, used incorrectly it will give incorrect results.

K_a=[H⁺]²/[ACID]

Once again - approximation only, in general it is incorrect.

And you can't combine them - first one can de derived from the second one using simplifying assumption that concentration of acid has not changed, thus they are contradicting.