[dds*90210]

Calculate the pH of a solution formed by mixing 20mL of 0.250M Formic Acid with 8mL of 0.400M NaOH
The ka of Formic Acid is 1.8 x 10^-4

so first i changed the concentration due to the dilution
then i set up my reaction. Na is a spectator ion
and then i set up my chart to solve for x and ultimatley, the pH

HCOOH + NaOH --> H2O + HCOO
0.179M   0.114M     ---     0
-x          -x            ---     +x
----------------------------------
0.179-x   0.114-x    ----    x

set this all equal to ka -->   ka = (x) / [( 0.179-x) (0.114-x)]
                                      x = 3.67 x 10^-6
                                      pOH = -log (x) , pH = poH - 14 = 13.06 <----thats my answer. but the book says the pH is 3.99


what was done wrong??

[Borek]

This is a buffer solution, that means using Henderson-Hasselbalch equation.

Note that your approach - apart from being wrong - is inconsistent with itself. If x is amount of NaOH that reacted, pOH = -log(0.114-3.67e-6).