Question

A 0.100 M solution of an acid, HA, has a pH = 2.00. What is the value of the ionization constant, Ka for this acid?

Answer 1

`Ka=1.11x10^(-3)`

Step-by-step explanation:

Hello,

In this case, since the ionization of the given HA acid is:

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

The equilibrium expression is:

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

Whereas the concentration of hydrogen ions is compute from the pH=

`[H^+]=10^(-pH)=10^(-2.00)=0.01M`

Which also equals the concentration of
`A^-` and the in general the ionization extent, therefore, the acid ionization constant, Ka, turns out:

`Ka=(0.01*0.01)/(0.1-0.01)\\ \\Ka=1.11x10^(-3)`

Regards.