2.45 mol HF is added to enough 0.102-M NaF solution to give a final volume of 2.1 L. What is the pH of the resulting solution given that the Ka of HF is 3.5x10-4 under these conditions?

pH = 2.40

Explanation:

It is possible to answer this question using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA]

For HF / F⁻ buffer, [HA] is weakacid, HF, and F⁻ is conjugate base, [A⁻].

pKa of this buffer is -log Ka → 3.456

As molarity of NaF is 0.102M and molarity of HF is: 2.45mol / 2.1L = 1.167M. pH of resulting solution is:

pH = 3.456 + log₁₀ [0.102] / [1.167]

pH = 2.40

I hope it helps!

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