A 1.36 L buffer solution consists of 0.185 M butanoic acid and 0.267 M sodium butanoate.

Question

A 1.36 L buffer solution consists of 0.185 M butanoic acid and 0.267 M sodium butanoate. Calculate the pH of the solution following the addition of 0.078 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The 𝐾a of butanoic acid is 1.52×10−5 . pH=

Answer 1

0.078 mol / 1.36 L = 0.057 M of NaOH

This will neutralize 0.057 M of butanoic acid and add 0.057 M to sodium benzoate.

Now you can make an ICE table with the new initial volumes:

0.185 - 0.057 = 0.128 M of butanoic acid

0.267 - 0.057 = 0.210 M of sodium benzoate (which will be benzoate in our equilibrium reaction because it dissociates into Na+ and C4H8O2-)

HC4H8O2 === H + C4H8O2

I 0.128 0 0.21

C -x +x +x

E 0.128 - x x 0.21 + x

1.52E-5 = x(0.21 + x) / (0.128 - x)

Now solve for x using the quadratic formula (after rearranging the equation algebraically)

x is the [H+]

pH = -log[H+]

Or you can use the henderson hasselbalch equation: pH = pKa + log(Base/Acid)

pKa = -log(Ka)

pH = -log(1.52 x 10^-5) + log(0.21/0.128) = 5.033

Check my math

pH =