What quantity in moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 3.90? (Ka for HF is 6.8 × 10⁻⁴)

Answer 1

For an acidic buffer, to calculate the pH, we can generally use the Henderson Hasselbalch equation:

pH = pKa + log [conjugate base] / [weak acid] = pKa + log [NaF] / [HF]

pKa = -log Ka = - log 6.8x10^-4 = 3.17

[HF] = 0.200 M

[NaF] = ?

Since we begin with 200.0 ml (0.2000 L) of 0.200 M HF, this is equal to 0.0400 mols HF

Plugging in values into the HH equation:

3.90 = 3.17 + log [NaF] / [HF]

log [NaF] / [HF] = 0.73

[NaF] / [HF] = 5.37

x / 0.04-x = 5.37

0.2148 - 5.37x = x

6.37x = 0.2148

x = 0.0337 moles NaF

HF + NaOH ==> NaF + H2O

0.04............x.................0...............Initial

-x................x................+0.0337....Change

0.04-.0337..-x................0.0337....Equilibrium

[HF] = 0.04 mols - 0.0337 mols = 0.0063 moles

[NaF] = 0.0337 moles = 0.0337 mols NaOH needed to be added (answer)



To check this, plug this back into HH equation and solve for pH. It should be 3.90.

pH = pKa + log [NaF] / [HF]

pH = 3.17 + log (0.0337 / 0.0063)

pH = 3.17 + 0.729

pH = 3.90