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:

K_{a} = ( [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. K_{a} = [(0.050) (0.050)] / 0.26K_{a} = 0.0096=9.6 *10^-3 (answer)

K_{a} = [(0.050) (0.050)] / 0.26K_{a} = 0.0096=9.6 *10^-3 (answer)

K_{a} = 0.0096

=9.6 *10^-3 (answer)

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