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-)#
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]#
In your case, the concentration of hydronium cations will be equal to
#["H"_3"O"^(+)] = color(red)(2) * "0.002 M"#
#["H"_3"O"^(+)] = "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.
