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

##### 1 Answer

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

...because only the

#"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

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

However, what you probably meant was the ** conjugate base** of

The

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

Thus...

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

And so, the *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