The #"pH"# of a #"0.08 mol dm"^(-3)# solution of #"HClO"# is equal to #2.85#. Calculate the ionization constant of the acid?

1 Answer
Jan 7, 2018

Answer:

#2.5 * 10^(-5)#

Explanation:

The idea here is that the #"pH"# of the solution will give you the equilibrium concentration of hydronium cations, which, in turn, will give you the equilibrium concentration of the hypochlorite anions and of the hypochlorous acid.

#"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "ClO"_ ((aq))^(-)#

Now, you know that when #1# mole of hypochlorous acid dissociates, it produces #1# mole of hypochlorite anions and #1# mole of hydronium cations.

This means that if the solution contains

#["H"_ 3"O"^(+)] = 10^(-"pH") quad "M"#

then it must also contain

#["ClO"^(-)] = 10^(-"pH") quad "M"#

and

#["HClO"] = (0.08 - 10^(-"pH")) quad "M"#

This is what you'd expect the equilibrium concentration of hypochlorous acid to be equal to if its dissociation produced #10^(-"pH")# #"M"# of hydronium cations.

In other words, in order for the dissociation of the acid to produce #["H"_3"O"^(+)] = 10^(-"pH")# #"M"#, the concentration of the acid must decrease by #["H"_3"O"^(+)] = 10^(-"pH")# #"M"#.

So now that you know the equilibrium concentrations for all three chemical species that are of interest here, you can plug them into the expression of the acid dissociation constant, #K_a#.

#K_a = (["ClO"^(-)] * ["H"_ 3"O"^(+)])/(["HClO"])#

You will end up with

#K_a = (10^(-"pH") * 10^(-"pH"))/(0.08 - 10^(-"pH"))#

which gets you

#K_a = (10^(-2.85))^2/(0.08 - 10^(-2.85)) = color(darkgreen)(ul(color(black)(2.5 * 10^(-5))))#

I'll leave the answer rounded to two sig figs, but keep in mind that you have only one significant figure for the initial concentration of the acid.