Add 0.40 g of NaOH to 50.0 mL of 0.10 M HCOOH, and find approximate pH. No volume change, Ka HCOOH=1.8x10^-4.

I get pH 8. Right?

I did not get 8. It appears to me that all of the HCOOH is neutralized and there is an excess of NaOH in 50 mL.