Question

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

Thanks in advance!

Answer 1

pKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5


`ka= ([H^+][CH3CH2COO^-])/([CH3CH2COOH])`


`[CH3CH2COO^-]`= 0.05

`[CH3CH2COOH]`= 0.10

Therefore 1.259x10^-5 =
`([H^+][0.05])/([0.1])`
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore
`[H^+] = ((1.259*10^-5)(0.1))/(0.05)`

Therefore
`[H^+]= 2.513*10^-5`

pH= -log [
`H^+`] = -log(2.513*10^-5)= 4.59.