Question 2bd63

Apr 13, 2016

$\text{pH} = 0.40$

Explanation:

You can calculate the pH of a solution of sulfuric acid by assuming that sulfuric acid acid behaves as a strong acid in both of its ionization reactions.

This implies that sulfuric acid, ${\text{H"_2"SO}}_{4}$, will donate both of its acidic protons to water to form sulfate anions, ${\text{SO}}_{4}^{2 -}$, and hydronium cations, ${\text{H"_3"O}}^{+}$. This means that every mole of sulfuric acid will produce two moles of hydronium cations in aqueous solution.

The first ionization looks like this

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

The first ionization is characteristic of a strong acid, i.e. the reaction will come very, very close to completion.

The second ionization reaction looks like this

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

This reaction does not go to completion, but the equilibrium still lies mostly to the right, so you can assume that it does go to completion.

Add the two ionization reactions to get

$\left\{\begin{matrix}{\text{H"_ 2"SO"_ (4(aq)) + "H"_ 2"O"_ ((l)) -> color(red)(cancel(color(black)("HSO"_ (4(aq))^(-)))) + "H"_ 3"O"_ ((aq))^(+) \\ color(red)(cancel(color(black)("HSO"_ (4(aq))^(-)))) + "H"_ 2"O"_ ((l)) rightleftharpoons "SO"_ (4(aq))^(2-) + "H"_ 3"O}}_{\left(a q\right)}^{+}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}$

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

So, if one mole of sulfuric acid produces $\textcolor{red}{2}$ moles of hydronium cations, it follows that the concentration of the hydronium cations will be twice that of the acid.

$\left[{\text{H"_3"O"^(+)] = color(red)(2) xx ["H"_2"SO}}_{4}\right]$

In your case, this will get you

["H"_3"O"^(+)] = 2 xx "0.20 M" = "0.40 M"

Now, the pH of the solution, which is simply a measure of how many hydronium cations you have present, is given by the equation

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#

Plug in the concentration of hydronium cations to get

$\text{pH} = - \log \left(0.40\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 0.40 \textcolor{w h i t e}{\frac{a}{a}} |}}}$