determine the [H3O],the pH and % ionization of a 0.1M HC2H3O2.given Ka=1.8x10-5?

Answer 1

Lancenigo di Villorba (TV), Italy

As you know, acetic acid (e.g. CH3COOH) is an organic compound showing acidic behaviours.
In aqueous media, the stuff undergoes to this chemical equilibrium :

CH3COOH(aq) <---> CH3COO-(aq) + H+(aq)

As you know, it exists a mathematical relation describing how chemical stuff's concentrations are bound at equilibrium's constant.
The close similarities binding acetic acid and its anion lies one lonely equilibrium's equation, as follows :

Ka = |CH3COO-| * |H+| / |CH3COOH|

as acidity's equilibrium of acetic acid.
These concentrations respect the conservation of matter, e.g. the acetic acid amount present in the aqueous medium.

|CH3COOH| + |CH3COO-| = C0

where C0 is nominal concentration of acetic acid.
Water also form hydrogen ions, as follows :

H2O(aq) <---> H+(aq) + OH-(aq)

When I refer to acidity's constant of acetic acid, I see that the latter is an acid stronger than water. In this manner, I will assume hydrogen ion and anion's acid have the same concentration value since they derived by same reaction and same stiochiometry.

YOUR CALCULUS
Before you apply the overwritten formula, you must determine the avalilable concentrations of acetic acid and its anion.
If I assume anion's concentration is similar to hydrogen ion's one hence I can get these simplifications :

x = |H+| = |CH3COO-|
C0 - x = |CH3COOH|

where C0 is nominal concnetration of acetic acid, e.g. 0.100 M.
The acidity's equilibrium becomes :

Ka = x * x / (C0 - x)

and this equation have solution as x = 1.3E-3 M.
|H+| = x = 1.3E-3 M
pH = 2.9

I hope this helps you

Answer 2

Rather more simply:

[H3O]+ = root (Ka x molarity]

pH = -log{H3O]+ and

% ionisation = root (Ka/molarity) x 100.

Answer 3

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