Question #8bf92
1 Answer
Explanation:
The thing to remember about sulfuric acid is that you can assume that it acts as a strong acid in both ionizations, not just in the first one.
"H"_ 2"SO"_ (4(aq)) + 2"H"_ 2"O"_ ((l)) -> color(red)(2)"H"_ 3"O"_ ((aq))^(+) + "SO"_ (4(aq))^(2-)H2SO4(aq)+2H2O(l)→2H3O+(aq)+SO2−4(aq)
This implies that, for all intended purposes, you can say that a solution of sulfuric acid will always contain
["H"_3"O"^(+)] = color(red)(2) * ["H"_2"SO"_4][H3O+]=2⋅[H2SO4]
In your case, the concentration of hydronium cations will be equal to
["H"_3"O"^(+)] = color(red)(2) * "0.002 M"[H3O+]=2⋅0.002 M
["H"_3"O"^(+)] = "0.004 M"[H3O+]=0.004 M
Now, the
color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))
This means that your solution will have
"pH" = - log(0.004)
= - log(4 * 10^(-3))
= - [log(4)+ log(10^(-3))]
= - log(4) - (-3) * overbrace(log(10))^(color(blue)(=1))
= 3 - log(4)
= color(darkgreen)(ul(color(black)(2.4)))
The answer is rounded to one decimal place, the number of sig figs you have for the concentration of the acid.
