Calculate the pH of a mixture of 100 cm3 of 0.500 mol dm–3 methanoic acid and 50 cm3 of 0.100 mol dm–3 sodium methanoate. (pKa methanoic acid = 3.75)

Answer 1

methanoic acid is a weak acid.

sodium methanoate is the salt of that weak acid (it is a conjugate base)

Together, they form a BUFFER.

To find the pH of a buffer, you can use the Henderson Hasselbalch equation which is...

pH = pKa + log [salt]/[acid]

We know the pKa = 3.75 and are asked to find the pH. So, we need to first find the [salt] and [acid]

moles salt = 50 cm3 x 1 L/1000 cm3 x 0.100 mol/L = 0.005 moles salt

moles acid = 100 cm3 x 1l/1000 cm2 x 0.500 mol/L = 0.05 moles acid

Total volume = 50 cm3 + 100 cm3 = 150 cm3 = 0.150 L

Final [salt] = 0.005 moles/0.15 L = 0.0333 M

Final [acid] = 0.05 moles/0.15 L = 0.333 M

pH = pKa + log [salt]/[acid]

pH = 3.75 + log (0.0333/0.333)

pH = 3.75 + (-1)

pH = 2.75

Answer 2

You can use the Hendersen-Haaselbach equation to solve this problem.

pH = pKa + log [A^-]/[HA]

A^- = sodium methanoate

HA = methanoic acid

Since the volumes are additive by combining both solutions, the final volume of the mixture = 150 cm³ (100 + 50). Need to calculate the final concentration of A^- and HA after mixing using the dilution equation.

[A^-] = 0.100 x 50/150 = 0.0333 mol dm^-3

[HA] = 0.500 x 100/150 = 0.333 mol dm^-3

pH = 3.75 + log (0.0333/0.333) = 3.75 - 1.00 = 2.75