Calculate the pH of a 0.10 M `H_3PO_4` solution that is also 0.0005 M `(NH_4)_3PO_4`?

Unfortunately, these `K_a` values are not real; the actual values are:

`K_(a1) = 7.5 xx 10^(-3)`
`K_(a2) = 6.3 xx 10^(-8)`
`K_(a3) = 4.5 xx 10^(-13)`

The magnitude of `K_(a3)` suggests that we can ignore it, so let's just focus on the first two.

When I do that, I get `"0.0274 M H"_3"O"^(+)` and `"pH" ~~ 1.57` when assuming that `"PO"_4^(3-)` reasonably associates in water to be mostly in the form of `"HPO"_4^(2-)`, forming the monohydrogen phosphate common ion.

Apparently, the salt was enough to negate the second dissociation.

DISCLAIMER: REALLY LONG ANSWER!

The `"0.0005 M"` `("NH"_4)_3"PO"_4` contributes the phosphate ion as the initial concentration. But we first have to find out how much `"HPO"_4^(2-)` it contributes as the common ion.

`"PO"_4^(3-)(aq) + "H"_2"O"(l) rightleftharpoons "HPO"_4^(2-)(aq) + "OH"^(-)(aq)`

`"I"" "0.0005" "" "" "" "-" "" "" "0" "" "" "" "0`
`"C"" "-x" "" "" "" "" "-" "" "+x" "" "" "+x`
`"E"" "0.0005-x" "" "-" "" "" "x" "" "" "" "x`

This has `K_(b1) = K_w/(K_(a3)) = (10^(-14))/(4.5 xx 10^(-13)) = 0.0222`, which is quite non-negligible, meaning that the reaction goes to near-completion.

`K_(b1)` is large compared to `0.0005`, so we assume that from the salt, we get `["HPO"_4^(2-)]_(eq) ~~ ["PO"_4^(3-)]_i = "0.0005 M"`.

(In actuality, it is `"0.000489 M"`, only `2.25%` error.)

The first... actual acid ICE table is:

`"H"_3"PO"_4(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "H"_2"PO"_4^(-)(aq)`

`"I"" "0.10" "" "" "" "-" "" "" "0" "" "" "" "" "0`
`"C"" "-x" "" "" "" "-" "" "+x" "" "" "" "+x`
`"E"" "0.10-x" "" "-" "" "" "x" "" "" "" "" "x`

The first `K_(a)` then gives:

`7.5 xx 10^(-3) = (["H"_3"O"^(+)]["H"_2"PO"_4^(-)])/(["H"_3"PO"_4])`

`= (x^2)/(0.10-x)`

The small `x` approximation would have given:

`= x^2/(0.10)`

`=> x = sqrt(0.10 cdot 7.5 xx 10^(-3)) = "0.0274 M"`

That is the first `["H"_3"O"^(+)]`, which we will use in the next dissociation:

`["H"_3"O"^(+)]_(eq1) ~~ "0.0274 M" = ["H"_3"O"^(+)]_(i2)`

`["H"_2"PO"_4^(-)]_(eq1) ~~ "0.0274 M" -= ["H"_2"PO"_4]_(i2)`

`["H"_3"PO"_4]_(eq1) ~~ "0.0726 M"`

So now, we have the second ICE table, already accounting for the `"PO"_4^(3-)` from `("NH"_4)_3"PO"_4` associating in water:

`"H"_2"PO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "HPO"_4^(2-)(aq)`

`"I"" "0.0274" "" "" "-" "" "" "0.0274" "" "" "0.0005`
`"C"" "-x" "" "" "" "-" "" "" "+x" "" "" "+x`
`"E"" "0.0274-x" "-" "" "" "0.0274+x" "0.0005+x`

Now, the cumulative `"H"_3"O"^(+)` would be the solution we would get for `["H"_3"O"^(+)]_(eq)` here, i.e.

`["H"_3"O"^(+)]_(eq) = ["H"_3"O"^(+)]_(eq1) + x`

Set up the second `K_a` expression:

`6.3 xx 10^(-8) = (["H"_3"O"^(+)]["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)])`

`= ((0.0274+x)(0.0005+x))/(0.0274-x)`

In this case, since `K_(a2)` `"<<"` `0.0274`, we can definitely use the small `x` approximation here. This gives:

`6.3 xx 10^(-8) = ((0.0274cancel(+x)^"small")(0.0005+x))/(0.0274cancel(-x)^"small")`

`~~ 0.0005 + x`

Therefore,

`x ~~ -"0.0005 M"`

Apparently, the salt is enough to reverse the direction of this reaction.

`color(blue)(["H"_3"O"^(+)]_(eq)) = ["H"_3"O"^(+)]_(eq1) + ["H"_3"O"^(+)]_(eq2)`

`= "0.0274 M" + (-"0.0005 M")`

`~~` `color(blue)("0.0269 M")`

So, the `"pH"` is:

`"pH" = -log["H"_3"O"^(+)]_(eq) = color(blue)(1.57)`

Following the same ICE tables as the answer above (? I will check to make sure. If they don't match, I will upload a picture of my work asap)

1. derived from table 1, `K_(a_1)=1.0xx10^-6=(x*x)/(0.10-x)`
   `x^2\cong(1.0xx10^-6)(0.10)`
   `x=\approx3.2xx10^-4` M
2. derived from table 2, `K_(a_2)=1.0xx10^-8=(x(0.00032+x))/(0.00032-x)`
   `x\cong((1.0xx10^-8)(\cancel(0.00032)))/(\cancel(0.00032))`
   `x\rArr1.0xx10^-8` M
3. table 3 with `K_(a_3)` calculates `x=0`, so negligible.

   Finding pH:
   `[H^+]_(\text(total))=0.0032+0.00032001=0.00064001` M
   which is `6.4xx10^-4` M of `[H^+]`
   pH is `-log[H^+]=-log(6.4xx10^-4)\approx3.19`