pH of 0.56M HF (pKa = 3.14)

[jokulsa]

How do I find the pH of 0.56M of hydrofluoric acid HF?

(pKa = 3.14)

[GrahamKemp]

Look to the left. At the top of the menu is "Chemistry tool". Click this. Then follow the link on the next page to "pH calculator".

In the text box labelled "Enter components of a solution to calculate pH" enter:

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HF pKa=3.14 c=0.56

Then click "Compute pH" and scroll down to the answer.

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Components of the solution (mixture):

Name: HF
Initial concentration after solutions are mixed = 0.56
Ka₁ = 0.00072443596007499 ( pKa₁ = 3.14 )

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Answer:

[H+] = 0.019782630261034
[OH-] = 5.0549395444635E-13
pH = 1.7037159659284
pOH = 12.296284034072
Ka₁ = 0.00072443596007499 ( pKa₁ = 3.14 )

[GrahamKemp]

Or by hand:

The equilibrium equation, where

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-log Ka=3.14

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Ka = [H][F]/[HF]

If there is no other source of hydronium ions than the acid (assuming a negligible contribution from water) then

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[H]=[F]

If the initial concentration of HF is

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C=0.56

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Ka = [H]{2} / (C-[H])

Which rearranges to:

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[H]= (-Ka + sqrt(Ka{2} + 4 C Ka ))/2

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[H]= (-10^{-3.14} + sqrt( 10^{-6.28 x 2} + 4 x 0.56 x 10^{-3.14} ))/2

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[H] = 0.01978264090522295443508279585463

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-log10 [H] = 1.7037157322531579844483334967812

[Reatele]

Probably no need to write out the solution for this question, right?

The pH of the 0.56 M HF solution is approximately 1.70.