If 48 g of butanoic acid C4H8O2 is dissolved in enough water

If 4.8 g of butanoic acid, C4H8O2, is dissolved in enough water to make 1.0 L of solution, what is the resulting pH? A table with Ka values can be found here.

Solution

Let HA represent butanoic acid

Initial concentration = mass/molar mass/volume

= 4.8/88.11/1.0 = 0.05448 M


.......HA...<=>...H+...+...A-

I..0.05448........0.........0

C....-a............+a........+a

E.0.05448-a......a.........a


Ka = [H+][A-]/[HA]

= a^2/(0.05448 - a) = 1.5 x 10^(-5)

a^2 + 1.5 x 10^(-5)a - 8.172 x 10(-7) = 0


The positive root of the quadratic equation is:

a = 8.965 x 10^(-4)


[H+] = a = 8.965 x 10^(-4) M

pH = -log[H+] = -log(8.965 x 10^(-4)) = 3.05