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Calculate the pH of a 0.10 M solution of hypochlorous acid, HOCl. K_a of HOCl is 3.5*10^-8?

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The ionization equilibrium equation for hypochlorous acid is:

HOCl(aq) <--> H+(aq) + OCl-(aq)

The acid constant equation is:

Ka = [H+(aq)][OCl-(aq)] / [HOCl(aq)]

in one liter of this solution, we have in the beginning 0.10 mole of hypochlorous acid, and no moles of hydrogen or hypochlorite ions. As the system reaches equilibrium, "x" molecules of hypochlorous acid ionized, leaving "0.10 - x" moles of unionized hypochlorous acid, and "x" moles of both hydrogen and hypochlorite ions. Substituting into the acid constant equation we get:

Ka = [H+(aq)][OCl-(aq)] / [HOCl(aq)]

3.5 x 10^-8 = [x][x] / [0.10 - x]

Given a very small acid ionization constant, very few molecules of hypochlorous acid ionized, leaving the concentration of unionized hypochlorous acid virtually unchanged. We can, therefore re-write the equation as follows:

3.5 x 10^-8 = [x][x] / [0.10]

3.5 x 10^-8 = [] / [0.10]

x^2 = (3.5 x 10^-8)(0.10)

x^2 = 35 x 10^-10

x = 5.9 x 10^-5 M = [H+]

pH = - log [H+]

pH = - (- 4.23)

pH = 4.23

Answer: The pH of a 0.10 M solution of hypochlorous acid is 4.23.

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HOCl = H+ + OCl- Initial concentration 0.010 M 0 0 At equilibrium 0.010(1 - a) 0.010a 0.010a Here 'a' is the degree of dissociation Ka = [H+][OCl-]/[HOCl] As Ka is small [HOCl] in the denominator can be neglected. Then Ka = [H+][OCl-] 3.5x10^-8 = 0.010a x 0.010a = 0.0001 a^2 = 10^-4 a^2 a^2 = 3.5x10^-8/10^-4 = 3.5x10^-4 a = 1.87 x 10^-2 Thus [H+] = 0.01 x 1.87 x 10^-2 = 1.87 x 10^-4 Hence pH = - log [H+] = - log (1.87 x 10^-4) = 3.728

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