Chemistry

Question

A buffer is made that is .10M propanoic acid (pKa = 4.9) and .05M sodium propanoate. Calculate the pH.

Thanks in advance!

1 Answer

  • 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.
NEWS TODAY