Calculate the pH of a 0.10 M $H_{3} P O_{4}$ $H_3PO_4$ solution that is also 0.0005 M ${(N H_{4})}_{3} P O_{4}$ $(NH_4)_3PO_4$ ?

Question

Calculate the pH of a 0.10 M $H_{3} P O_{4}$ $H_3PO_4$ solution that is also 0.0005 M ${(N H_{4})}_{3} P O_{4}$ $(NH_4)_3PO_4$ ?

$K_{a_{1}} = 1.0 \times 10^{- 6}$ $K_(a_1)=1.0xx10^-6$
$K_{a_{2}} = 1.0 \times 10^{- 8}$ $K_(a_2)=1.0xx10^-8$
$K_{a_{3}} = 1.0 \times 10^{- 10}$ $K_(a_3)=1.0xx10^-10$

Reactions used:

$H_{3} P O_{4} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + H_{2} P O_{4}^{-} (a q)$ $H_3PO_4(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+H_2PO_4^(-)(aq)$

$H_{2} P O_{4}^{-} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + H P O_{4}^{2 -} (a q)$ $H_2PO_4^(-)(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+HPO_4^(2-)(aq)$

$H P O_{4}^{2 -} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + P O_{4}^{3 -} (a q)$ $HPO_4^(2-)(aq)+H_2O(l)\rightleftharpoonsH_3O^+(aq)+PO_4^(3-)(aq)$

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

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Answer 1 · 2018-06-27T20:39:20

Unfortunately, these $K_{a}$ $K_a$ values are not real; the actual values are:

$K_{a 1} = 7.5 \times 10^{- 3}$ $K_(a1) = 7.5 xx 10^(-3)$
$K_{a 2} = 6.3 \times 10^{- 8}$ $K_(a2) = 6.3 xx 10^(-8)$
$K_{a 3} = 4.5 \times 10^{- 13}$ $K_(a3) = 4.5 xx 10^(-13)$

The magnitude of $K_{a 3}$ $K_(a3)$ suggests that we can ignore it, so let's just focus on the first two.

When I do that, I get ${0.0274 M H}_{3} O^{+}$ $"0.0274 M H"_3"O"^(+)$ and $pH \approx 1.57$ $"pH" ~~ 1.57$ when assuming that ${PO}_{4}^{3 -}$ $"PO"_4^(3-)$ reasonably associates in water to be mostly in the form of ${HPO}_{4}^{2 -}$ $"HPO"_4^(2-)$ , forming the monohydrogen phosphate common ion.

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

DISCLAIMER: REALLY LONG ANSWER!

The $0.0005 M$ $"0.0005 M"$ ${({NH}_{4})}_{3} {PO}_{4}$ $("NH"_4)_3"PO"_4$ contributes the phosphate ion as the initial concentration. But we first have to find out how much ${HPO}_{4}^{2 -}$ $"HPO"_4^(2-)$ it contributes as the common ion.

${PO}_{4}^{3 -} (a q) + H_{2} O (l) ⇌ {HPO}_{4}^{2 -} (a q) + {OH}^{-} (a q)$ $"PO"_4^(3-)(aq) + "H"_2"O"(l) rightleftharpoons "HPO"_4^(2-)(aq) + "OH"^(-)(aq)$

$I 0.0005 - 00$ $"I"" "0.0005" "" "" "" "-" "" "" "0" "" "" "" "0$
$C - x - + x + x$ $"C"" "-x" "" "" "" "" "-" "" "+x" "" "" "+x$
$E 0.0005 - x - x x$ $"E"" "0.0005-x" "" "-" "" "" "x" "" "" "" "x$

This has $K_{b 1} = \frac{K_{w}}{K_{a 3}} = \frac{10^{- 14}}{4.5 \times 10^{- 13}} = 0.0222$ $K_(b1) = K_w/(K_(a3)) = (10^(-14))/(4.5 xx 10^(-13)) = 0.0222$ , which is quite non-negligible, meaning that the reaction goes to near-completion.

$K_{b 1}$ $K_(b1)$ is large compared to $0.0005$ $0.0005$ , so we assume that from the salt, we get ${[{HPO}_{4}^{2 -}]}_{e q} \approx {[{PO}_{4}^{3 -}]}_{i} = 0.0005 M$ $["HPO"_4^(2-)]_(eq) ~~ ["PO"_4^(3-)]_i = "0.0005 M"$ .

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

The first... actual acid ICE table is:

$H_{3} {PO}_{4} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + H_{2} {PO}_{4}^{-} (a q)$ $"H"_3"PO"_4(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "H"_2"PO"_4^(-)(aq)$

$I 0.10 - 00$ $"I"" "0.10" "" "" "" "-" "" "" "0" "" "" "" "" "0$
$C - x - + x + x$ $"C"" "-x" "" "" "" "-" "" "+x" "" "" "" "+x$
$E 0.10 - x - x x$ $"E"" "0.10-x" "" "-" "" "" "x" "" "" "" "" "x$

The first $K_{a}$ $K_(a)$ then gives:

$7.5 \times 10^{- 3} = \frac{[H_{3} O^{+}] [H_{2} {PO}_{4}^{-}]}{[H_{3} {PO}_{4}]}$ $7.5 xx 10^(-3) = (["H"_3"O"^(+)]["H"_2"PO"_4^(-)])/(["H"_3"PO"_4])$

$= \frac{x^{2}}{0.10 - x}$ $= (x^2)/(0.10-x)$

The small $x$ $x$ approximation would have given:

$= \frac{x^{2}}{0.10}$ $= x^2/(0.10)$

$\Rightarrow x = \sqrt{0.10 \cdot 7.5 \times 10^{- 3}} = 0.0274 M$ $=> x = sqrt(0.10 cdot 7.5 xx 10^(-3)) = "0.0274 M"$

That is the first $[H_{3} O^{+}]$ $["H"_3"O"^(+)]$ , which we will use in the next dissociation:

${[H_{3} O^{+}]}_{e q 1}^{+} \approx 0.0274 M = {[H_{3} O^{+}]}_{i 2}^{+}$ $["H"_3"O"^(+)]_(eq1) ~~ "0.0274 M" = ["H"_3"O"^(+)]_(i2)$

${[H_{2} {PO}_{4}^{-}]}_{e q 1} \approx 0.0274 M \equiv {[H_{2} {PO}_{4}]}_{i 2}$ $["H"_2"PO"_4^(-)]_(eq1) ~~ "0.0274 M" -= ["H"_2"PO"_4]_(i2)$

${[H_{3} {PO}_{4}]}_{e q 1} \approx 0.0726 M$ $["H"_3"PO"_4]_(eq1) ~~ "0.0726 M"$

So now, we have the second ICE table, already accounting for the ${PO}_{4}^{3 -}$ $"PO"_4^(3-)$ from ${({NH}_{4})}_{3} {PO}_{4}$ $("NH"_4)_3"PO"_4$ associating in water:

$H_{2} {PO}_{4}^{-} (a q) + H_{2} O (l) ⇌ H_{3} O^{+} (a q) + {HPO}_{4}^{2 -} (a q)$ $"H"_2"PO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "HPO"_4^(2-)(aq)$

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

Now, the cumulative $H_{3} O^{+}$ $"H"_3"O"^(+)$ would be the solution we would get for ${[H_{3} O^{+}]}_{e q}^{+}$ $["H"_3"O"^(+)]_(eq)$ here, i.e.

${[H_{3} O^{+}]}_{e q}^{+} = {[H_{3} O^{+}]}_{e q 1}^{+} + x$ $["H"_3"O"^(+)]_(eq) = ["H"_3"O"^(+)]_(eq1) + x$

Set up the second $K_{a}$ $K_a$ expression:

$6.3 \times 10^{- 8} = \frac{[H_{3} O^{+}] [{HPO}_{4}^{2 -}]}{[H_{2} {PO}_{4}^{-}]}$ $6.3 xx 10^(-8) = (["H"_3"O"^(+)]["HPO"_4^(2-)])/(["H"_2"PO"_4^(-)])$

$= \frac{(0.0274 + x) (0.0005 + x)}{0.0274 - x}$ $= ((0.0274+x)(0.0005+x))/(0.0274-x)$

In this case, since $K_{a 2}$ $K_(a2)$ $<<$ $"<<"$ $0.0274$ $0.0274$ , we can definitely use the small $x$ $x$ 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:

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

Answer 2 · 2018-07-03T05:18:19

This is using the $K_a$ 's given by the problem. Just wanted to check my work.

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 $-log[H^+]=-log(6.4xx10^-4)\approx3.19$

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Calculate the pH of a 0.10 M $H_{3} P O_{4}$ $H_3PO_4$ solution that is also 0.0005 M ${(N H_{4})}_{3} P O_{4}$ $(NH_4)_3PO_4$ ?

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