Question

The weak acid HA is 4 % ionized (dissociated) in a 0.30 M solution. A. What is Ka for this acid?B. What is the pH of this solution?

Answer 1

`Ka=5x10^(-4)`

`pH=1.92`

Step-by-step explanation:

Hello,

In this case, given the percent ionization and the concentration of the acid, one computes the concentration of hydrogen ions as follows:

`\% ionization=([H^+])/([HA])*100\%`

`[H^+]=(\% ionization*[HA])/(100\%) =(4\%*0.30M)/(100\%)=0.012M`

Therefore the Ka is computed by using the equilibrium expression:

`Ka=([H^+][A^-])/([HA]) =(0.012M*0.012M)/(0.30M-0.012M)\\ \\Ka=5x10^(-4)`

And the pH:

`pH=-log([H^+])=-log(0.012)\\\\pH=1.92`

Regards.