Question

I can't solve this! I keep trying but I can't get it.

Suppose 25mL of 1.00M HCl is added to 50mL of 1.00M HF. What is the pH of the HF in this solution?

The pH is supposed to be 0.48

Thanks.

-K.

Answer 1

You need to be aware that the HF is a weak acid and ionizes only partially with a Ka of about 7.2 x 10^-4 so the (H^+) contributed by the HF, in comparison to the 100% ionized HCl, is too small to count it. So the pH of the solution is essentially the pH of the HCl.

You have diluted 1.00 M x 25/75 = 0.333 M HCl
and pH = -log 0.333 = 0.477 which rounds to 0.48.