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

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#K_(a_1)=1.0xx10^-6#

#K_(a_2)=1.0xx10^-8#

#K_(a_3)=1.0xx10^-10#

Reactions used:

#H_3PO_4(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+H_2PO_4^(-)(aq)#
#H_2PO_4^(-)(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+HPO_4^(2-)(aq)#
#HPO_4^(2-)(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+PO_4^(3-)(aq)#

We are asked to use an ICE table at least once.

Reactions used:

#H_3PO_4(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+H_2PO_4^(-)(aq)# #H_2PO_4^(-)(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+HPO_4^(2-)(aq)# #HPO_4^(2-)(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+PO_4^(3-)(aq)#

We are asked to use an ICE table at least once.

##### 2 Answers

Unfortunately, these

#K_(a1) = 7.5 xx 10^(-3)#

#K_(a2) = 6.3 xx 10^(-8)#

#K_(a3) = 4.5 xx 10^(-13)#

The magnitude of

When I do that, I get

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

**DISCLAIMER:** *REALLY LONG ANSWER!*

The **phosphate ion** as the initial concentration. But we first have to find out how much **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

(In actuality, it is

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

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

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

The small

#= x^2/(0.10)#

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

That is the first

#["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

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

Set up the second

#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

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

#### Answer:

This is using the

#### Explanation:

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)

- 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 - 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 - 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#