Calculate the #["H"_3"O"^(+)]# and #["OH"^(-)]# for 0.0051 M H2SO4. Answer in units of M.?

#["H"_3"O"^(+)] = "0.0081 M"#; why is it NOT #"0.0102 M"#?

#["OH"^(-)] = K_w/(["H"_3"O"^(+)]) = ???#

Begin by recognizing that #"H"_2"SO"_4# is a strong acid that dissociates completely into the #"HSO"_4^(-)# and #"H"_3"O"^(+)#.

#"H"_2"SO"_4(aq) + "H"_2"O"(l) -> "HSO"_4^(-)(aq) + "H"_3"O"^(+)(aq)#

Thus, #["H"_3"O"^(+)]# from THE FIRST dissociation is #"0.0051 M"#. Are we done? I hope not.

What about the second dissociation? No acid dissociation constant #K_(a2)# was provided, and so I must use Google to GUESS that #K_(a2) ~~ 1.2 xx 10^(-2)#. The equilibrium is written

#"HSO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "SO"_4^(2-)(aq) + "H"_3"O"^(+)(aq)#

#"I"" "["HSO"_4^(-)]_i" "" "-" "" "" "0" "" "" "" "["H"_3"O"^(+)]_i#
#"C"" "-x" "" "" "" "-" "" "+x" "" "" "" "+x#
#"E"" "["HSO"_4^(-)]_i - xcolor(white)(.)-" "" "" "x" "" "" "["H"_3"O"^(+)]_i+x#

and the mass action expression is known to be:

#1.2 xx 10^(-2) = (["SO"_4^(2-)]["H"_3"O"^(+)])/(["HSO"_4^(-)]) -= (x(["H"_3"O"^(+)]_i + x))/(["HSO"_4^(-)]_i - x)#

We must identify that

• #["H"_2"SO"_4]_i = ["HSO"_4^(-)]_(eq) = ["H"_3"O"^(+)]_(eq)# from the first dissociation is #["HSO"_4^(-)]_i# and #["H"_3"O"^(+)]_i# for the second dissociation.
• #["HSO"_4^(-)]_(eq) = ["H"_3"O"^(+)]_(eq)# for the first dissociation.
• #["SO"_4^(2-)]_(eq) ne ["H"_3"O"^(+)]_(eq)# for the second cumulative dissociation.

This #K_(a2)# is not small, so we must solve the full quadratic.

#1.2 xx 10^(-2) = (x(0.0051 + x))/(0.0051 - x)#

#=> x^2 + 0.0171x + 6.12 xx 10^(-5) = 0#

Solving this gives #x = "0.00304 M"# for #["H"_3"O"^(+)]# for THE SECOND dissociation. Therefore,

#color(blue)(["H"_3"O"^(+)]) = "0.0051 M" + "0.00304 M" = color(blue)("0.0081 M")#

And from here, the #K_w# of water gives the #["OH"^(-)]# at #25^@ "C"#.

#color(blue)(["OH"^(-)]) = K_w/(["H"_3"O"^(+)]) = 10^(-14)/("0.0081 M")#

#= color(blue)(1.24 xx 10^(-12) "M")#