[ThatGuy]

Find the pH of a solution formed by dissolving 0.100 moles of HC₂H₃O₂ with a Ka of 1.8 x 10^-5 and 0.200 moles of NaC₂H₃O_2(s) in a total volume of 2.5 L.

Chemical Equation:

CH₃COOH_(aq) + NaC₂H₃O_2(s) ↔ Na⁺(aq) + CH₃COO^-_(aq) + H⁺ _(aq)

Initial Concentration of CH₃COOH_(aq) : 0.100 moles / 2.5 L = 0.040 M

Initial Concentration of NaC₂H₃O_2(s) : 0.200 moles / 2.5 L = 0.080 M

Ka = 1.8 x 10^-5 = ((x)(2x)²(x)) / ((0.040)(0.080))
 
x = 0.01095 M

pH = -log (0.01095) = 1.96

[Hunter2]

Use Henderson Hasselbalch Equation for this.

[ThatGuy]

I haven't learned the Henderson Hasselbalch Equation yet, although I have googled it.

Is it wrong to approach the problem in the manner I initially did?

[mafagafo]

HH will make it easier. It will not be the only way though.
As Borek (I may be mistaken) pointed to me a while ago, HH is derived from the definition of K.

Firstly, the concentration of solids is not taken into account. See what you did?

Both sodium and acetate usually make salts with high solubility in water. NaC₂H₃O₂ is really soluble.
As everything here is an approximation, let's say it is 100% soluble.

Lastly, write Ka symbolically. What is Ka? Does it take [Na⁺] into account?

What is you equilibrium equation now?

What is your new answer?

[Borek]

Is it wrong to approach the problem in the manner I initially did?

I am afraid yes, it is off. You wrote some random equation, you assumed for no apparent reason that its equilibrium constant will be that of acetic acid dissociation, then you plugged in some numbers and an unknown, which you for no apparent reason assume will describe concentrations of all substances present (as if they were guaranteed to be equal). This is pretty random, it can't yield a correct result.

mafagafo is right, HH equation is a rearranged dissociation constant, just written in a way that makes pH calculation of some solutions pretty easy. It will work here perfectly well.