Question

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

Answer 1

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

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

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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")`