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#