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

## ${K}_{{a}_{1}} = 1.0 \times {10}^{-} 6$ ${K}_{{a}_{2}} = 1.0 \times {10}^{-} 8$ ${K}_{{a}_{3}} = 1.0 \times {10}^{-} 10$ Reactions used: ${H}_{3} P {O}_{4} \left(a q\right) + {H}_{2} O \left(l\right) \setminus r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} \left(a q\right) + {H}_{2} P {O}_{4}^{-} \left(a q\right)$ ${H}_{2} P {O}_{4}^{-} \left(a q\right) + {H}_{2} O \left(l\right) \setminus r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} \left(a q\right) + H P {O}_{4}^{2 -} \left(a q\right)$ $H P {O}_{4}^{2 -} \left(a q\right) + {H}_{2} O \left(l\right) \setminus r i g h t \le f t h a r p \infty n s {H}_{3} {O}^{+} \left(a q\right) + P {O}_{4}^{3 -} \left(a q\right)$ We are asked to use an ICE table at least once.

Jun 27, 2018

Unfortunately, these ${K}_{a}$ values are not real; the actual values are:

${K}_{a 1} = 7.5 \times {10}^{- 3}$
${K}_{a 2} = 6.3 \times {10}^{- 8}$
${K}_{a 3} = 4.5 \times {10}^{- 13}$

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

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

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

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

${\text{PO"_4^(3-)(aq) + "H"_2"O"(l) rightleftharpoons "HPO"_4^(2-)(aq) + "OH}}^{-} \left(a q\right)$

$\text{I"" "0.0005" "" "" "" "-" "" "" "0" "" "" "" } 0$
$\text{C"" "-x" "" "" "" "" "-" "" "+x" "" "" } + x$
$\text{E"" "0.0005-x" "" "-" "" "" "x" "" "" "" } x$

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

${K}_{b 1}$ 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 $\text{0.000489 M}$, only 2.25% error.)

The first... actual acid ICE table is:

${\text{H"_3"PO"_4(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "H"_2"PO}}_{4}^{-} \left(a q\right)$

$\text{I"" "0.10" "" "" "" "-" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-x" "" "" "" "-" "" "+x" "" "" "" } + x$
$\text{E"" "0.10-x" "" "-" "" "" "x" "" "" "" "" } x$

The first ${K}_{a}$ then gives:

$7.5 \times {10}^{- 3} = \left(\left[{\text{H"_3"O"^(+)]["H"_2"PO"_4^(-)])/(["H"_3"PO}}_{4}\right]\right)$

$= \frac{{x}^{2}}{0.10 - x}$

The small $x$ approximation would have given:

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

$\implies x = \sqrt{0.10 \cdot 7.5 \times {10}^{- 3}} = \text{0.0274 M}$

That is the first $\left[{\text{H"_3"O}}^{+}\right]$, which we will use in the next dissociation:

${\left[{\text{H"_3"O"^(+)]_(eq1) ~~ "0.0274 M" = ["H"_3"O}}^{+}\right]}_{i 2}$

${\left[{\text{H"_2"PO"_4^(-)]_(eq1) ~~ "0.0274 M" -= ["H"_2"PO}}_{4}\right]}_{i 2}$

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

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

${\text{H"_2"PO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "HPO}}_{4}^{2 -} \left(a q\right)$

$\text{I"" "0.0274" "" "" "-" "" "" "0.0274" "" "" } 0.0005$
$\text{C"" "-x" "" "" "" "-" "" "" "+x" "" "" } + x$
$\text{E"" "0.0274-x" "-" "" "" "0.0274+x" } 0.0005 + x$

Now, the cumulative ${\text{H"_3"O}}^{+}$ would be the solution we would get for ${\left[{\text{H"_3"O}}^{+}\right]}_{e q}$ here, i.e.

${\left[{\text{H"_3"O"^(+)]_(eq) = ["H"_3"O}}^{+}\right]}_{e q 1} + x$

Set up the second ${K}_{a}$ expression:

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

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

In this case, since ${K}_{a 2}$ $\text{<<}$ $0.0274$, 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")

$\approx 0.0005 + x$

Therefore,

$x \approx - \text{0.0005 M}$

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")

$\approx$ $\textcolor{b l u e}{\text{0.0269 M}}$

So, the $\text{pH}$ is:

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

Jul 3, 2018

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)

1. derived from table 1, ${K}_{{a}_{1}} = 1.0 \times {10}^{-} 6 = \frac{x \cdot x}{0.10 - x}$
${x}^{2} \setminus \cong \left(1.0 \times {10}^{-} 6\right) \left(0.10\right)$
$x = \setminus \approx 3.2 \times {10}^{-} 4$ M
2. derived from table 2, ${K}_{{a}_{2}} = 1.0 \times {10}^{-} 8 = \frac{x \left(0.00032 + x\right)}{0.00032 - x}$
$x \setminus \cong \frac{\left(1.0 \times {10}^{-} 8\right) \left(\setminus \cancel{0.00032}\right)}{\setminus \cancel{0.00032}}$
$x \setminus \Rightarrow 1.0 \times {10}^{-} 8$ M
3. table 3 with ${K}_{{a}_{3}}$ calculates $x = 0$, so negligible.

Finding pH:
${\left[{H}^{+}\right]}_{\setminus \textrm{\to t a l}} = 0.0032 + 0.00032001 = 0.00064001$ M
which is $6.4 \times {10}^{-} 4$ M of $\left[{H}^{+}\right]$
pH is $- \log \left[{H}^{+}\right] = - \log \left(6.4 \times {10}^{-} 4\right) \setminus \approx 3.19$