GS-1 Physical Sciences Question 14

[dnpgr16513]

Hi,

So if this acid was considered to be a "weak acid" would we then proceed with the Ka equation?

For example, if H2SO3 was weak ( which it isn't) would we calculate it like this?

Ka= [H30+][SO32-]/ [HSSO3]

5 X 10^-6 = (X)^2/.01

Is this correct? Would we assume ( hypothetical question if this was weak) that the concentration of H30 and SO32- were the same?

Thanks so much!!!!!!!!!!!!!!!!!!

[jellywing_2058]

That is correct; if the acid was a weak acid, you would have to use the ICE method (or IRF table) to calculate the [H+] and then calculate the pH.

However, you do not assume that the concentrations of both [H3O+] and [SO32-] are the same. In fact, you need to calculate the concentrations of both as they are considered x-variables in the ICE table.

You essentially have the initial concentrations of H2SO3 (0.01 M), H3O+ (0 M) and SO32- (0 M). The change would be +x for both H3O+ and SO32-; therefore, at equilibrium, the concentrations would be (0.01 – x) M for each H3O+ and SO32-.

You can then calculate the results using the ICE table in which Ka = [H30+][SO32-] / [H2SO3] where you substitute the Ka and ICE table variables and then you calculate the [H30+] to then go to the pH calculation.