if the equilibrium constantof the reaction of weak acid HA with strong base is 10^9, then the pH of 0.1M NaA solution wlii be?

Ka=[H+(aq)][A−(aq)][HA(aq)]

Here, we can now make two assumptions which will grately simplify the calculations;

Assumption One: [H⁺(aq)] = [A^-(aq)]
This means that we are ignoring the effect of the ionisation of water, this is a valid assumption for most 'normal' concentrations.
Assumption Two: As weak acids dissociate very little, we can assume that the concentration of the acid at equilibrium is the same as if the acid did not dissociate

With these assumptions we can rewrite the equilibrium constant as follows;
Ka=[H+(aq)]2[HA(aq)] 

Ka=[H+(aq)]2[HA(aq)]

Although water is not considered ionic, it does dissociate to some extent. Water is amphoretic (able to act both as a base and an acid) and will auto-dissociate into the oxonium ion H₃O⁺ and the hydroxide ion;

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