Question

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

Answer 1

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

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...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?