Suppose that you have 125 mL of a buffer that is 0.140 M in both hydrofluoric acid ( HF ) and its conjugate base ( F − ) . Calculate the maximum volume of 0.370 M HCl that can be added to the buffer before its buffering capacity is lost.

38,7mL of HCl 0,370M

Explanation:

The buffering capacity is lost when the pH of the solution is out of pka ± 1.

For the buffer:

HF ⇄ H⁺ + F⁻; pka = 3,20

Moles of both HF and F⁻ are:

HF: 0,140M ₓ 0,125L = 0,0175 moles of HF

Moles of F⁻ are the same.

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ F⁻ / HF

pH = 3,20 + log F⁻ / HF (1)

The addition of X moles of HCl (H⁺) shift the equilibrium to the left, thus:

         HF             ⇄   H⁺  +     F⁻

0,0175 moles + X             0,0175 moles - X          

The buffer lost its capacity when pH = 3,20 - 1 = 2,20. Thus, replacing in (1):

2,20 = 3,20 + log ₁₀0,0175 - X / 0,0175 + X

0,1 = 0,0175 - X / 0,0175 + X

0,00175 + 0,1X = 0,0175 - X

-0,01575 = -1,1X

0,0143 moles = X

You need to add 0,0143 moles of HCl. In volume for a concentration of 0,370M HCl:

0,0143 moles ÷ 0,370M = 0,0387L of HCl 0,370M ≡ 38,7mL of HCl 0,370M

I hope it helps!

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