Calculate the ph of a 0.400 m nach3co2 solution. ka for acetic acid, ch3co2h, is 1.8×10-5

We can break this down.
First, we have a salt called sodium acetate. This ionic salt will completely dissociate into 0.400 M concentrations of Na+ and CH3COO- ions. Sodium won't react in any appreciably acidic way with the water, so we can ignore the sodium ions.
The acetate ions on the other hand are the conjugate base of the weak acid called acetic acid. This means the acetate ions will extract an acidic proton from the water, leaving behind OH- in the solution, making it basic. 
A weak base will establish an equilibrium with the water with which it reacts, which will have the form CH3COO- + H2O ⇌ CH3COOH + OH-
Remembering that a K value is equal to the product concentration over the reactants, and that we can exclude water as a pure liquid from the equilibrium equation, we get a final equation of: Kb = [CH3COOH]*[OH-]/[CH3COO-]
Where [molecule] implies the concentration.
Even though we aren't given the Kb value of the solution, we can calculate it from the equation: Kb * Ka(conjugate acid) = Kw (10^-14)
We know that the Ka of the conjugate acid is 1.8*10^-5, so we can plug it into the equation to find that:
Kb = (10^-14)/(1.8*10^-5)
Kb = 5.6 * 10^-10
This makes our final values:
[CH3COOH] = x
[OH-] = x
[CH3COO-] = 0.400 - x (we can say that x is too small to affect 0.400 in any way, so we'll treat [CH3COO-] as though it remains 0.400 when we plug it in)
Kb = 5.6 * 10^-10
Therefore:
Kb = [CH3COOH]*[OH-]/[CH3COO-]
5.6*10^-10 = (x * x)/(0.400)
2.2 * 10^-10 = x^2
x = 1.5 * 10^-5
The concentration of OH- ions in the solution is 1.5 * 10^-5 M
We're not done yet, but we're so close.
We know that the pOH (not pH) is the -log([OH-]), and that pOH + pH = pKw(14)
pOH = -log(1.5*10^-5) = 4.83
pOH + pH = 14
4.83 + pH = 14
pH = 9.17, meaning the solution is basic
This makes sense, as we added a basic salt to the solution, so the final solution should be basic.

Sorry i dont understand the question can you rephase it?