A 0.01 M solution of an acid has a pH = 5.0. What is the Ka value?

I'm assuming that you're working with a monoprotic weak acid here so that the ionization equilibrium can be written like this

#"HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "A"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#

Now, you know that the solution has

#"pH" = 5#

As you know, the #"pH"# of the solution is defined as

#color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))#

You can rearrange this equation to find the equilibrium concentration of hydronium cations.

#log(["H"_3"O"^(+)]) = - "pH"#

This is equivalent to

#10^log(["H"_3"O"^(+)]) = 10^(-"pH")#

which gets you

#["H"_3"O"^(+)] = 10^(-"pH")#

In your case, you will have

#["H"_3"O"^(+)] = 10^(-5.0) = 1.0 * 10^(-5)# #"M"#

Now, notice that every mole of #"HA"# that ionizes produces #1# mole of #"A"^(-)#, the conjugate base of the acid, and #1# mole of hydronium cations.

This means that, at equilibrium, the solution has

#["A"^(-)] = ["H"_3"O"^(+)] " "-># produced in a #1:1# mole ratio

In your case, you have

#["A"^(-)] = 1.0 * 10^(-5)# #"M"#

The initial concentration of the acid will decrease because some of the molecules ionize to produce #"A"^(-)# and #"H"_3"O"^(+)#.

So, in order for the ionization to produce #["H"_3"O"^(+)]#, the initial concentration of the acid must decrease by #["H"_3"O"^(+)]#.

This means that, at equilibrium, the concentration of the weak acid will be equal to

#["HA"] = ["HA"]_"initial" - ["H"_3"O"^(+)]#

In your case, you will have

#["HA"] = "0.01 M" - 1.0 * 10^(-5)color(white)(.)"M"#

#["HA"] = "0.00999 M"#

By definition, the acid dissociation constant, #K_a#, will be equal to

#K_a = (["A"^(-)] * ["H"_3"O"^(+)])/(["HA"])#

Plug in your values to find

#K_a = (1.0 * 10^(-5)color(red)(cancel(color(black)("M"))) * 1.0 * 10^(-5)color(white)(.)"M")/(0.00999color(red)(cancel(color(black)("M"))))#

#K_a = 1.001 * 10^(-8)# #"M"#

Rounded to one significant figure and expresses without added units, the answer will be

#color(darkgreen)(ul(color(black)(K_a = 1 * 10^(-8))))#