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 4.00? (Ka for HF is 6.8 × 10⁻⁴)

Answer 1

When you add NaOH to HF, you will create a buffer composed of the weak acid (HF) and the conjugate base (F^-) as indicated below:

HF + OH^- ==> F^- + H₂O

We can then use the Henderson Hasselbalch equation to find the amount of OH^- needed:

pH = pKa + log [conj.base]/[acid]

pKa = -log Ka = 3.17

4.00 = 3.17 + log [F^-]/[HF]

log [F^-]/[HF] = 0.833

[F^-]/[HF] = 6.81

Looking at an ICE table (initial moles HF = 0.200 mol/L x 0.2 L = 0.0400 moles)

HF + OH^- ==> F^-

0.04.....x............0.....I

-x.......-x..........+x....C

0.04-x...0.........x.....E

Substituting in [F^-]/[HF] = 6.81, we have...

x/0.04-x = 6.81

x = 0.035 = moles of F^- are needed

Since the reaction is 1 mole NaOH to make 1 mole F^-, we would need to add 0.035 moles NaOH

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You can check this as follows:

Final [F^-] = 0.035 mols / 0.2 L = 0.175 M

Final [HF] = 0.04 mols - 0.035 mols = 0.005 mols/0.2 L = 0.025 M

pH = 3.17 + log (0.175/0.025)

pH = 3.17 + log 7

pH = 3.17 + 0.845

pH = 4.01