What is the Ka of the acid?

Question

Harriet dissolves 1.0 mole of a weak monoprotic acid in 2.0 liters of water. She measures the concentration of H30+ ions to be 3.4 X 10-2 M.

Answer 1

The initial concentration of the acid is C=mole/V=1/2=0.5 mole/L
Let's assume that the acid is HA

.. .. .. .. .. .. .. HA <=> H(+) + A(-)
Initial .. .. .. .. 0.5
Dissociate .. .x
Produce .. .. .. .. .. .. .. x .. .. .. x
At Equil. .. . 0.5-x .. .. . x .. .. .. x

Ka=[H+][A-]/[HA] =x^2/(0.5-x)

but x=[H+]=3.4*10^-2 so

Ka= ((3.4*10^-2)^2) / (0.5-3.4*10^-2) =2.48*10^-3