What is the $K_{b}$ $K_b$ for $H S O_{4}^{-}$ $HSO_4^(-)$ if its $K_{a}$ $K_a$ is $1.99$ $1.99$ ?

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What is the $K_{b}$ $K_b$ for $H S O_{4}^{-}$ $HSO_4^(-)$ if its $K_{a}$ $K_a$ is $1.99$ $1.99$ ?

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Answer 1 · 2017-08-02T07:31:52

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

...because only the $K_{a 1}$ $K_(a1)$ of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ is related to the $K_{b}$ $K_b$ for ${HSO}_{4}^{-}$ $"HSO"_4^(-)$ , and only the $K_{a}$ $K_a$ of ${HSO}_{4}^{-}$ $"HSO"_4^(-)$ (i.e. the $K_{a 2}$ $K_(a2)$ of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ ) is related to the $K_{b}$ $K_b$ of ${SO}_{4}^{2 -}$ $"SO"_4^(2-)$ .

$H_{2} {SO}_{4} K_{a 1} ⇌ {HSO}_{4}^{-} K_{a 2} ⇌ {SO}_{4}^{2 -}$ $"H"_2"SO"_4 stackrel(K_(a1)" ")(rightleftharpoons) "HSO"_4^(-) stackrel(K_(a2)" ")(rightleftharpoons) "SO"_4^(2-)$
$_{b 1}^{K_{b 1}}_{b 2}^{K_{b 2}}$ $" "" "" "^(K_(b1))" "" "" "" "^(K_(b2))$

Answering the question as-written would require looking up the $K_{a 1}$ $K_(a1)$ of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ , which is $\approx$ $~~$ $1000$ $1000$ .

Recall that

$K_{a} K_{b} = K_{w} = 10^{- 14}$ $K_aK_b = K_w = 10^(-14)$ ,

${pK}_{a} + {pK}_{b} = {pK}_{w} = 14$ $"pK"_a + "pK"_b = "pK"_w = 14$ ,

at $25^{\circ} C$ $25^@ "C"$ and $1 atm$ $"1 atm"$ .

So, the " $K_{b}$ $K_b$ of ${HSO}_{4}^{-}$ $"HSO"_4^(-)$ " (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?

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What is the $K_{b}$ $K_b$ for $H S O_{4}^{-}$ $HSO_4^(-)$ if its $K_{a}$ $K_a$ is $1.99$ $1.99$ ?

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What is the $K_{b}$ $K_b$ for $H S O_{4}^{-}$ $HSO_4^(-)$ if its $K_{a}$ $K_a$ is $1.99$ $1.99$ ?