Question #2bd63
1 Answer
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,
The first ionization looks like this
#"H"_ 2"SO"_ (4(aq)) + "H"_ 2"O"_ ((l)) -> "HSO"_ (4(aq))^(-) + "H"_ 3"O"_((aq))^(+)#
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
#"HSO"_ (4(aq))^(-) + "H"_ 2"O"_ ((l)) rightleftharpoons "SO"_ (4(aq))^(2-) + "H"_ 3"O"_((aq))^(+)#
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
#{("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"_((aq))^(+)) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)#
#"H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> "SO"_ (4(aq))^(2-) + color(red)(2)"H"_ 3"O"_((aq))^(+)#
So, if one mole of sulfuric acid produces
#["H"_3"O"^(+)] = color(red)(2) xx ["H"_2"SO"_4]#
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
#"pH" = - log(0.40) = color(green)(|bar(ul(color(white)(a/a)0.40color(white)(a/a)|)))#