Question

When 0.285 g of a solid weak acid HA (molar mass = 184 g mol−1) is dissolved in water to a total volume of 25.0 mL, the pH of the solution is 3.59.

What is the acid ionization constant (Ka) of the acid?

Answer 1

will it be Ka=HA/gxvol?

Answer 2

I don't think it's quite that simple.
(HA) = mols/L = g/molar mass/L = 0.285/184/0.025 = about 0.06.
................HA ==> H^+ + A^-
I...............0.06.......0..........0
C...............-x..........x...........x
E............0.06-x......x............x

The problem tells you pH = 3.59 so 3.59 = -log(H^+) and (H^+) = 2.57E-4 M.
Plug the E line into the Ka expression and you get (x)(x)/(0.06-x) = Ka.
You know what x is; i.e., 2.57E-4 so plug that in and solve for Ka.