jaidencyrocki6734 jaidencyrocki6734
  • 03-05-2018
  • Chemistry
contestada

The ka of hf is 6.8 x 10-4. what is the ph of a 0.35 m solution of hf?

Respuesta :

superman1987
superman1987 superman1987
  • 12-05-2018
when the reaction equation is:

HF ↔ H+   +   F-

and when the Ka expression
= concentration of products/concentration of reactions

so, Ka = [H+][F-]/[HF]

when we assume:

[H+] = [F-] = X

and [HF] = 0.35 - X

So, by substitution:

6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X

∴ X = 0.015 M

∴[H+] = X = 0.015

when PH = -㏒[H+]

∴PH = -㏒0.015

        = 1.8
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