The Calculation of pKa in a Chemical Experiment

Calculating pKa in a Chemical Experiment

During an experiment, a student mixed 14.5 ml of a 0.500 M sodium fluoride solution with 15.6 ml of a 0.750 M hydrofluoric acid solution. The measured pH was 3.24. Now, we need to calculate the pKa of the hydrofluoric acid used in the experiment.

The pKa of hydrofluoric acid is 3.44.

To solve this problem, we use Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]) where pH is the measured pH, pKa is the dissociation constant of the acid, [A⁻] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

In this case, the acid is hydrofluoric acid (HF), which can donate a proton (H⁺) to form the conjugate base, fluoride ion (F⁻).

The concentration of the fluoride ion can be calculated from the sodium fluoride solution that was added:

moles of F- = 0.500 mol/L x 0.0145 L = 0.00725 mol

concentration of F- = 0.00725 mol / (0.0145 L + 0.0156 L) = 0.234 M

The concentration of the hydrofluoric acid can be calculated in a similar way:

moles of HF = 0.750 mol/L x 0.0156 L = 0.0117 mol

concentration of HF = 0.0117 mol / (0.0145 L + 0.0156 L) = 0.375 M

Now we substitute these values into Henderson-Hasselbalch equation:

3.24 = pKa + log([0.234]/[0.375])

Solving for pKa, we get:

pKa = 3.24 - log([0.234]/[0.375])

pKa = 3.24 - log(0.624)

pKa = 3.24 + 0.204

pKa = 3.44