What is the #K_b# for #HSO_4^(-)# if its #K_a# is #1.99#?

1 Answer
Aug 2, 2017

This question can't quite be answered in its exact wording...


...because only the #K_(a1)# of #"H"_2"SO"_4# is related to the #K_b# for #"HSO"_4^(-)#, and only the #K_a# of #"HSO"_4^(-)# (i.e. the #K_(a2)# of #"H"_2"SO"_4#) is related to the #K_b# of #"SO"_4^(2-)#.

#"H"_2"SO"_4 stackrel(K_(a1)" ")(rightleftharpoons) "HSO"_4^(-) stackrel(K_(a2)" ")(rightleftharpoons) "SO"_4^(2-)#
#" "" "" "^(K_(b1))" "" "" "" "^(K_(b2))#

Answering the question as-written would require looking up the #K_(a1)# of #"H"_2"SO"_4#, which is #~~# #1000#.

Recall that

#K_aK_b = K_w = 10^(-14)#,

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

at #25^@ "C"# and #"1 atm"#.

So, the "#K_b# of #"HSO"_4^(-)#" (as you have stated) is...

#K_(b1) = K_w/K_(a1) = 10^(-14)/(10^3) ~~ ul(10^(-11))#

However, what you probably meant was the #K_b# of #ul("SO"_4^(2-))#, the conjugate base of #"HSO"_4^(-)#... but another issue is that I don't believe your #K_a#.

The #K_a# of #"HSO"_4^(-)# is about #0.012#, so you have supplied the #"pK"_a#, i.e. you have given

#"pK"_a = -log(K_a)#

Thus...

#10^(-"pK"_a) = K_a = 10^(-1.99) = 0.0102# is your #K_a#.

And so, the #K_b# for #"SO"_4^(2-)# (which is probably what you actually thought you wanted) is...

#color(blue)(K_(b2)) = K_w/K_(a2)#

#= 10^(-14)/0.0102#

#= ul(color(blue)(9.77 xx 10^(-13)))#

With a #K_b# this small, is #"HSO"_4^(-)# primarily an acid that shall dissociate, or a base that shall associate? i.e. will it be easier to form sulfate, or sulfuric acid in aqueous solution?