Below is the question that I am working on. Will someone check my work for part (a) and (b). I am also in need of some help on parts (c) and (d). The questions I Have are below.

The question is: A person mixes 50.0ml of 0.115 M HAc with 25.0 ml of 0.182 M NaOH and then dilutes the resulting mixture to 100.0ml.

a. Write the net ionic equation-----

**HC2H3O2(aq) + NaOH **

NaC2H3O2(aq) + H20[/b]

b. Determine which reagent is in excess. I determined that it is the

**HC2H3O2 that is in excess and the NaOH that is the limiting reagent. ** HC2H3O2=0.00575mol

NaOH= 0.00455mol

c. Calculate the concentration of HC2H3O2, C2H3O2 and Na in the final solution.

[NaAc] [H2o] -
- [/u]

[HAc] [NaOH] = [ 0.0455-x] [0.0455-x] =

I know that the Ka for HC2H3O2 is 1.8x10^-5

I am not sure how to figure out the number that I should make the above equation that I figutred out equal to. If someone could explain this to me that would be great.

d. Calculate the concentration H+ (aq) in this solution.

I know that in this problem I am dealing with:

H+ C2H3O2 Na OH and H2O.

The H+ from the C2H3O2 would be the only major contributor of H+ ions to the initial solution because the H2O acts either as a weak acid or weak base. These are my thoughts on this problem so far. I am not exactly sure of where to go from here.

I am thinking that maybe I should set up the problem where

[H+] [C2H3O2] = Ka for HC2H3O2 Ka =1.8*10^-5

HC2H3O2

Does this make sense to you? Then I would calculate it out for x where H+ equals x.

Thank you for your time and *delete me*!!

It is very much appreciated!