Buffer Titrations and pka

Question

A buffer solution containing the acid HX and its conjugate base X^-, has a pH=6.00.

A 10.00 mL sample required 9.50 mL 0.1000 M NaOH to titrate the sample to the Basic equivalence point.

13.50 mL 0.1000 M HCl were required to titrate a second 10.00 mL buffer sample to the Acidic equivalence point.

What is the pK_a for the acid HX?

I know that ph=pka + log([A^-]/[HA]), just am unsure how to utilize this.

Thanks:)

Answer 1

So, what you need to do here is to find the [A^-] and [HA], then plug that into the Henderson Hasselbalch equation and solve for pKa.

To find HA:

mols NaOH = 9.5 ml x 1 L / 1000 ml x 0.1000 mol/L = 0.00095 mols NaOH = 0.00095 mols HA

[HA] = 0.00095 mols / 10.00 ml x 1000 ml / L = 0.095 M

To find A^-:

mols HCl = 13.50 ml x 1 L / 1000 ml x 0.1000 mol/L = 0.00135 mols HCl = 0.00135 mols A^-

[A^-] = 0.00135 mols / 10.00 ml x 1000 ml / L = 0.135 M

HH equation:

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

6.00 = pKa + log (0.135/0.095) = pKa + 0.15

pKa = 6.00 - 0.15

pKa = 5.85