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

Answer 1

It's difficult to respond to this question exactly in its 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#.

Remember 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)#

So...

#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?
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Answer 2

[ K_b = \frac{K_w}{K_a} ] [ K_b = \frac{1.0 \times 10^{-14}}{1.99} ] [ K_b \approx 5.03 \times 10^{-15} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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