What is the Ka of Hydrofluoric acid at 35 ͦ C, if a 0.26M hydrofluoric acid solution has a pH of 1.30 at 35 ͦ C?
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Write the dissociation equation for the weak acid HF :
HF <===> H+ + F¯
2) Write the equilibrium expression:
Ka = ( [H+] [F¯] ) / [HF]
3).a) We will use the pH to calculate the [H+]. We know pH = -log [H+], therefore [H+] = 10¯pH
[H+] = 10¯1.3 = 0.050 M.
b) From the dissociation equation, we know there is a 1:1 molar ratio between [H+] and [F¯]. Therefore:[F¯] = 0.050 M
[F¯] = 0.050 M
c) the final value, [HF] is given in the problem. Ka = [(0.050) (0.050)] / 0.26Ka = 0.0096=9.6 *10^-3 (answer)
Ka = [(0.050) (0.050)] / 0.26Ka = 0.0096=9.6 *10^-3 (answer)
Ka = 0.0096
=9.6 *10^-3 (answer)
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