Hyrdrofluoric acid (Ka = 7.2 x 10 -4 ), is used in a wide range of commercial pr

Question

Hyrdrofluoric acid (Ka = 7.2 x 10-4), is used in a wide range of commercial processes. HF is used for glass etching and the synthesis of pharmaceutical drugs and compounds such as teflon. One issue with HF is its corrosive nature, "Aqueous hydrofluoric acid is a contact-poison with the potential for deep, initially painless burns and ensuing tissue death." (from Wikepedia page on hydrofluoric acid).

The flouride ion is often used in dental procedures, as the F- can react with tooth enamel to form highly stable mineral fluorohydroxyapatite (which is less likely to break down in the mouth than "normal" enamel). One common form of F- delivery is fluoride varnish, which has a F- concentration of around 1.778 M.

Given what you know of the acid base chemistry of HF, what is the concentration of HF in an aqueous solution with a pH of 5.85? (For retake exam, use the following format: 1.23 x 10-4should be entered as 1.23e-4)

Explanation / Answer

Answer – Given, [F-] = 1.778 , pH = 5.58 , Ka = 7.2*10-4

We know,

pH + pOH = 14

so, pOH = 14 – pH

              = 14 – 5.58

              = 8.42

So, [OH-] = 10-pOH

                 = 10-8.42

                 = 3.80*10-9 M

So we know reaction

   F- + H2O <------> HF + OH-

I 1.778                       0          0

C   -x                            +x       +x

E 1.778-x                     +x      3.80*10-9 M

Kb = 1*10-14 / 7.2*10-4

      = 1.39*10-11

So, Kb = [HF] [OH-] / [F-]

[OH-] =x = 3.80*10-9 M

[HF] = 1.778-x

        = 1.778 –x

        = 1.778

1.39*10-11 = x * 3.80*10-9 M / (1.778)

So, 3.80*10-9 x = 1.39*10-11 * 1.778

                         = 2.47*10-11

x = 2.47*10-11 / 3.80*10-9

    = 0.00650 M

[HF] = 0.00650 M

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