pKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5
[tex]ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]} [/tex]
[tex][CH3CH2COO^-]
[/tex]= 0.05
[tex][CH3CH2COOH][/tex]= 0.10
Therefore 1.259x10^-5 = [tex] \frac{[H^+][0.05]}{[0.1]} [/tex]
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore [tex][H^+] = \frac{(1.259*10^-5)(0.1)}{0.05} [/tex]
Therefore [tex][H^+]= 2.513*10^-5[/tex]
pH= -log [[tex]H^+[/tex]] = -log(2.513*10^-5)= 4.59.
