Question

What is the `"pH"` for a solution containing `"0.500 M"` of a weak triprotic acid whose first `K_a` is `5.78 xx 10^(-10)`? Assume only the first proton is significant. Report your answer to two significant figures.

Answer 1

The ICE table is the typical one for monoprotic weak acids, since we only look at the first proton here:

`"H"_3"A"(aq) + "H"_2"O"(l) rightleftharpoons "H"_2"A"^(-)(aq) + "H"_3"O"^(+)(aq)`

`"I"" "0.500" "" "" "-" "" "" "0" "" "" "" "" "0`
`"C"" "-x" "" "" "-" "" "+x" "" "" "" "+x`
`"E"" "0.500-x" "-" "" "" "x" "" "" "" "" "x`

The mass action expression is:

`K_(a1) = (["H"_2"A"^(-)]["H"_3"O"^(+)])/(["H"_3"A"]) = 5.78 xx 10^(-10)`

`= x^2/(0.500 - x)`

This `K_(a1)` is small, so to a great approximation, `x=["H"^(+)]` is indeed small in comparison to `"0.500 M"`. Thus, we make the approximation so that

`5.78 xx 10^(-10) ~~ x^2/0.500`

and so:

`["H"^(+)] = sqrt(5.78 xx 10^(-10) cdot 0.500) " M" = 1.7 xx 10^(-5)` `"M"`

and the `"pH"` would just be:

`"pH" = -log["H"^(+)] = 4.769`

To two sig figs it would then be:

`color(blue)("pH" = 4.8)`