- We've already talked about how to write an equilibrium expression, so
if we have some generic acid HA that donates a proton to H2O, H2O becomes H3O+ and HA turns into the conjugate base, which is A-. Here's our equilibrium expression and the ionization constant Ka for a weak acid. We already talked about the
fact that it's going to be less than one. Here we have three weak
acids: hydrofluoric acid, acetic acid, and methanol. Over here are the Ka values. You can see that hydrofluoric
acid has the largest Ka value, so even though
they're all considered to be weak acids, 3.5
times 10 to the negative 4 is larger than 1.8 times
10 to the negative 5. So hydrofluoric acid is
stronger than acetic acid, and acetic acid is stronger than methanol. But again, they're all
considered to be weak acids, relative to the stronger ones. Let's talk about pKas. The pKa is defined as the negative log of the Ka. If we wanted to find the pKa
for methanol, all we have to do is take the Ka and
take the negative log of it. So the pKa is equal to the negative log of 2.9 times 10 to the negative 16. Let's get out the calculator
and let's do that. Negative log of 2.9 times 10 to the negative 16. This gives us 15.54 when we round that. So the pKa of methanol is equal to 15.54. We could write in a pKa column right here, and for methanol it's 15.54. If you did the same
calculation for acetic acid, you would get 4.74; and
once again, if you did this for hydrofluoric acid, you would get 3.46. So as we go up on our table
here, we're increasing in acid strength. Out of our three weak
acids, hydrofluoric acid is the strongest, so it
has the largest value for Ka, but notice it has the
smallest value for the pKa. The lower the value for pKa,
the more acidic your acid. 3.46 is lower than 4.74,
and so hydrofluoric acid is more acidic than acetic acid.