[no username inserted]

In order to calculate the pH of a dilute HAB salt , in which HA is a weak acid and B is a weak base , shall I solve a 4th degree polynomial? Is there any simpler way to get the answer?
I already know that if the salt is not dilute , pH = 1/2 (pKa + pKaB) would come in use.
Thanks.

[Borek]

In order to calculate the pH of a dilute HAB salt , in which HA is a weak acid and B is a weak base , shall I solve a 4th degree polynomial? Is there any simpler way to get the answer?

How diluted? Once it is diluted enough, pH is just 7 

There is no simple answer to such a question, as a lot depends on relative strengths of acid, base and concentration. In some cases it can be possible to find approximations that will yield simpler solutions. But in general case the polynomial is the only sure way to go.

[no username inserted]

In order to calculate the pH of a dilute HAB salt , in which HA is a weak acid and B is a weak base , shall I solve a 4th degree polynomial? Is there any simpler way to get the answer?

How diluted? Once it is diluted enough, pH is just 7 

There is no simple answer to such a question, as a lot depends on relative strengths of acid, base and concentration. In some cases it can be possible to find approximations that will yield simpler solutions. But in general case the polynomial is the only sure way to go.

For instance , a solution of 10^-5 M of NH4CN. It's not diluted enough  What approximations can I use in this example?

[Borek]

For instance , a solution of 10^-5 M of NH4CN. It's not diluted enough  What approximations can I use in this example?

None as far as I am aware. Perhaps with pH above 9 you can ignore Kw, which will yield an easier to solve 3^rd degree polynomial.