What is the initial concentration of the conjugate base of a weak monoprotic acid whose #"pK"_a# is #4.30# if the #"pH"# is #8.30#?

Well, you gotta work backwards, and recognize what a conjugate base is (i.e. that it uses #K_b#, not #K_a#).

I got #["A"^(-)]_i = "0.0200 M"#.

The #"pH"# is taken at equilibrium, and thus,

#["H"^(+)]_(eq) = 10^(-"pH") = 10^(-8.30)#

#= 5.011 xx 10^(-9) "M H"^(+)#

Now, we know that the base associates in aqueous solution. In units of #"M"# we construct the ICE table:

#"A"^(-)(aq) + "H"_2"O"(l) rightleftharpoons "HA"(aq) + "OH""^(-)(aq)#

#"I"" "["A"^(-)]_i" "" "color(white)(.)-" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "-" "" "+x" "" "" "+x#
#"E"" "["A"^(-)]_i-xcolor(white)(.)-" "" "" "x" "" "" "" "x#

From the #"pK"_a# then we note that we are looking at a BASE! Therefore, we must use the #K_b#!

#"pK"_a + "pK"_b = "pK"_w = 14#

As a result, #"pK"_b = 14 - 4.30 = 9.70# for #"A"^(-)# from the #"pK"_a# of #"HA"#, and...

#K_b = 10^(-"pK"_b) = 10^(-9.70) = 1.995 xx 10^(-10)#

#= (["HA"]["OH"^(-)])/(["A"^(-)])#

#= x^2/(["A"^(-)]_i - x)#

And from #["H"^(+)]_(eq)# we just got earlier, we can find #["OH"^(-)]_(eq)#, i.e. our value of #x#...

At #25^@ "C"# and #"1 atm"#, the autoionization of water gives:

#K_w = ["H"^(+)]["OH"^(-)] = 10^(-14) = (5.011 xx 10^(-9) "M")["OH"^(-)]#

Thus,

#["OH"^(-)]_(eq) = x = 10^(-14)/(5.011 xx 10^(-9) "M")#

#= 1.995 xx 10^(-6) "M OH"^(-)#

And so:

#K_b = 1.995 xx 10^(-10)#

#= (1.995 xx 10^(-6) "M")^2/(["A"^(-)]_i - 1.995 xx 10^(-6) "M")#

#= (3.980 xx 10^(-12) "M"^2)/(["A"^(-)]_i - 1.995 xx 10^(-6) "M")#

Solve for #["A"^(-)]_i#.

#1.995 xx 10^(-10)(["A"^(-)]_i - 1.995 xx 10^(-6) "M") = 3.980 xx 10^(-12) "M"^2#

#1.995 xx 10^(-10)["A"^(-)]_i " M"^2 - 3.980 xx 10^(-16) " M"^2 = 3.980 xx 10^(-12) "M"^2#

#ul(["A"^(-)]_i) = (3.980 xx 10^(-12) "M"^2 + 3.980 xx 10^(-16) " M"^2)/(1.995 xx 10^(-10) "M")#

#= ul"0.01995 M A"^(-)#

or #color(blue)("0.0200 M A"^(-))# to three sig figs.