Question

The pH of 0.259 M solution of an unknown acid is measured to be 2.353. What is the Ka of the acid?

Answer 1

`Ka=7.73x10^(-5)`

Step-by-step explanation:

Hello!

In this case, since the pH is measure of the amount of hydrogen ions in solution, first given the pH we can compute the concentration of hydrogen ions via:

`pH=-log([H^+])`

`[H^+]=10^(-pH)=10^(-2.353)=4.436x10^(-3)M`

Thus, since the ionization of the given acid, leads to the following equilibrium expression as it is a weak one:

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

We notice the concentration of hydrogen ions equal the concentration of the acids conjugate base, so we can compute:

`Ka=(4.436x10^(-3)*4.436x10^(-3))/(0.259-4.436x10^(-3))\\\\Ka=7.73x10^(-5)`

Best regards!