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
The
#"pK"_a = -log(K_a)#
Thus...
#10^(-"pK"_a) = K_a = 10^(-1.99) = 0.0102# is your#K_a# .
And so, the
#color(blue)(K_(b2)) = K_w/K_(a2)#
#= 10^(-14)/0.0102#
#= ul(color(blue)(9.77 xx 10^(-13)))#
With a