Using Ka for a diprotic acid e.g. H2A --> 2H+ + A2- Would the expression be Ka = [H+]^2 * [A2-] / [H2A]? very confused about this I know a similar thing is done with Kc
Using Ka for a diprotic acid e.g. H2A --> 2H+ + A2- Would the expression be Ka = [H+]^2 * [A2-] / [H2A]? very confused about this I know a similar thing is done with Kc
No
Kc I'd different With kc, the formula is products over reactants to the power of the number of moles
With ka the formula is always - [H+] x [A-] / [HA]
I know at the half- neutralisation point A- = HA so pH= pKa but why is this only for weak acids or does it also apply for strong acids??, in the indicators section of the text book it says "the colour change of most indicators takes place over a pH range of around 2 units, centred around the value of pka for the indicator" but it doesnt talk about weak acids in this context