[Borek]

Mass balance - you are dissolving ammonium formate, at first for every HCOO^- there is one NH₄⁺ - so their concentrations are identical. Later some of the NH₄⁺ react with OH^- giving off ammonia - one molecule per each ion. Does sum of their concentrations change?

[bioguy2]

no, the sums don't change i don't think

[bioguy2]

so for the mass balance i got
[NH4] = [HCOO-] + [H2COO]

[Borek]

no, the sums don't change i don't think

Correct.

so for the mass balance i got
[NH4] = [HCOO-] + [H2COO]

That's a little bit off - you forgot about the ammonia. You did OK for the formate and formic acid, however, at that high pH (10) you can safely assume concentration of HCOOH to be neglectable (you may use the same approach you used for [NH₄⁺]/[NH₃] to check if that's OK).

At this stage you should be able to write three equations in three unknown - and solve.

[bioguy2]

i don't know why but i can't figure this out.

i have 5.69x10^-10 = [NH3(OH)][1.0x10^-10] / NH4+
   

1.757 x 10^-5 = [H2COO-][OH-] / [HCOO-]

[bioguy2]

haha sweet thanks i figured it out 2.2
thanks alot

[Borek]

Looks OK.