# Question 044a6

Jan 7, 2018

["H"_3"O"^(+)] > "0.25 M" $\to$ by a small amount.

#### Explanation:

I'm not really sure why you're supposed to "predict" what the concentration of hydronium cations will be if you are given the acid dissociation constant for the second ionization, but I'd say that you can predict that this solution will have

["H"_3"O"^(+)] > "0.25 M"

but only by a very small amount.

This is the case because most of the hydronium cations present in the solution will be produced by the first ionization since sulfuric acid is a weak acid in its second ionization.

So you can say that you have

["H"_ 3"O"^(+)]_ "total" = ["H"_ 3"O"^(+)]_ "1st ionization" + ["H"_ 3"O"^(+)]_ "2nd ionization"

$\textcolor{w h i t e}{\frac{a}{a}}$

So, to find the concentration of hydronium cations, you use the fact that sulfuric is actually a strong acid only in its first ionization, for which

${K}_{a 1} \text{ >> } 1$

This tells you that sulfuric acid ionizes completely in its first ionization to produce bisulfate anions, ${\text{HSO}}_{4}^{-}$, and hydronium cations, ${\text{H"_3"O}}^{+}$.

In its second ionization, sulfuric acid is no longer considered a strong acid, since

${K}_{a 2} < 1$

This tells you that the bisulfate anions, ${\text{HSO}}_{4}^{-}$, does not ionize completely to produce sulfate anions, ${\text{SO}}_{4}^{2 -}$, and hydronium cations.

So you can say that for the first ionization, you have

${\text{H"_ 2"SO"_ (4(aq)) + "H"_ 2"O"_ ((l)) -> "HSO"_ (4(aq))^(-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

Since the acid ionizes completely in $1 : 1$ mole ratios, you can say that for the first ionization of the acid, you have

$\left[{\text{HSO"_ 4^(-)] = ["H"_ 3"O"^(+)] = ["H"_ 2"SO}}_{4}\right]$

In your case, you will have

["HSO"_4^(-)] = ["H"_3"O"^(+)] = "0.25 M"

Now, when the bisulfate anion dissociates, it only does so partially, meaning that an equilibrium is established between the undissociated bisulfate anions and the sulfate anions and the hydronium cations.

${\text{HSO"_ (4(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "SO"_ (4(aq))^(2-) + "H"_ 3"O}}_{\left(a q\right)}^{+}$

If you take $x$ $\text{M}$ to be the concentration of bisulfate anions that dissociate to produce sulfate anions and hydronium cations, you can say that the resulting solution will contain

["SO"_4^(2-)] = x quad "M"

["H"_ 3"O"^(+)] = (x + 0.25)quad "M"

This is the case because the solution already contains $\text{0.25 M}$ of hydronium cations from the first ionization of the acid!

["HSO"_4^(-)] = (0.25 - x)quad "M"

This basically means that in order for the reaction to produce $x$ $\text{M}$ of sulfate anions and $x$ $\text{M}$ of hydronium cations, it must consume $x$ $\text{M}$ of bisulfate anions.

By definition, the acid dissociation constant for this second ionization is equal to

${K}_{a 2} = \left(\left[{\text{SO"_4^(2-)] * ["H"_ 3"O"^(+)])/(["HSO}}_{4}^{-}\right]\right)$

$0.012 = \frac{x \cdot \left(0.25 + x\right)}{0.25 - x}$

Rearrange to get a quadratic equation form

${x}^{2} + \left(0.25 + 0.012\right) \cdot x - 0.003 = 0$

This quadratic equation will produce two solutions, one positive and one negative. Since $x$ represents concentration, you can discard the negative solution and say that

$x = 0.01099$

This means that, at equilibrium, the solution will have

["H"_3"O"^(+)] = (0.25 + 0.01099)quad "M"

color(darkgreen)(ul(color(black)(["H"_3"O"^(+)] = "0.26 M")))

The answer is rounded to two decimal places, the number of decimal places you have for the concentration of sulfuric acid.

As predicted, you have

["H"_3"O"^(+)] = "0.26 M" > "0.25 M"#