What is the percent ionization in a 0.15 M solution of HF? (is it possible to calculate percent ionization with only molarity?)

No, you need the value of the acid dissociation constant, ##K_a##, for hydrofluoric acid.

If you do get the value of the acid dissociation constant, then you can definitely calculate the percent ionization.

Here's how that calculation would look like - I'll choose ##K_a = 7.2 * 10^(-4)##.

Start with the balanced chemical equation and use the ICE table method (more here: of the dissociated ions in solution

....##HF_((aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + F_((aq))^(-)##I...0.15M.................................0....................0C..(-x).....................................(+x).................(+x)E..(0.15-x)...............................x.....................x

According to the definition of the acid dissociation constant, you'll get

##(x * x)/(0.15 - x) = x^(2)/(0.15 - x) = 7.2 * 10^(-4)##

Solving for ##x## will get you ##x = 0.010039##. This is the concentration of both the hydronium, and of the fluride ions in solution.

Percent ionization is calculated by taking the concentration of hydronium ions, dividing it by the initial concentration of the acid and multiplying the ratio by 100.

##"% ionization" = "0.010039 M"/"0.15 M" * 100 = "6.7%"##

The answer can only have two , the number of sig figs given for 0.15M.