Question

consider an exceptionally weak acid, HA, with Ka= 1 x 10-20. you make 0.1M solution of the salt NA. what is the pH.

Answer 1

`pH=10.5`

Step-by-step explanation:

Hello,

In this case, the dissociation of the given weak acid is:

`HA\rightleftharpoons H^++A^-`

Therefore, the law of mass action for it turns out:

`Ka=([H^+][A^-])/([HA])`

That in terms of the change
`x` due to the reaction extent is:

`1x10^(-23)=(x*x)/(0.1-x)`

Thus, by solving with the quadratic equation or solver, we obtain:

`x=31.6x10^(-12)M`

Which clearly matches with the hydrogen concentration in the solution, therefore, the pH is:

`pH=-log(-31.6x10^(-12))\\pH=10.5`

Regards.