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

User Dmitry D
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1 Answer

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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
User Chetan Shirke
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