Calculate the ph of a solution that is 0.065 m in potassium propionate (c2h5cook or kc3h5o2) and 0.090 m in propionic acid (c2h5cooh or hc3h5o2).

When propionic acid is [HA] = 0.09 m

potassium propionate is [A-] = 0.065 m

and we have Ka = 1.3 x 10^-5

So we can use H-H equation:

PH = Pka + ㏒ [A-/HA]

when Pka = - ㏒ Ka

∴Pka = - ㏒ (1.3 x 10^-5)

= 4.89

by substitution:

∴PH = 4.89 + ㏒[0.065 / 0.09]

= 4.75