Question

What is the pH of a 0.384 M aqueous solution of hydrofluoric acid? K_a (HF) = 7.1x10^-4

Answer 1

The answer to this question is quite simple

We need to know 2 formulas

1

Ka = [H⁺] [F^-] / [HF]

Since concentration Of [H⁺] and [F^-] is same

So

Ka = [H⁺]² / [ HF]

[ H⁺] = √ Ka × [HF]

[H⁺] = √ (0.384× 7.1×10^-4)

[H⁺] = √ 0.00027264

= 0.016511814

Now we know

pH = -log[H+]

So

pH = -log [0.016511814]

= 1.78

So the pH of solution is

1.78

I hope this helps. If you have any query or want more detailed explanation, feel free to ask in the comments section below.