# Ionic equilibrium problem...?proof needed....

## Concentration of $\left[{H}^{+}\right]$ ion in a mixture of two weak acids... if ${\alpha}_{1}$ and ${\alpha}_{2}$ are their degree of dissociation which are negligible w.r.t unity...${K}_{{a}_{1}}$ and ${K}_{{a}_{2}}$ are ionization constant of the acids respectively and ${C}_{1}$ and ${C}_{2}$ is their concentration...

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Nov 10, 2017

$\left[{\text{H}}^{+}\right] = \frac{1}{2} \left(\sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{{K}_{a 1} {C}_{1} + 4 {K}_{a 2} {C}_{2}}\right)$

Note that this works best if ${K}_{a 1}$ and ${K}_{a 2}$ are both on the order of ${10}^{- 5}$ or less, i.e. the acids are sufficiently weak.

DISCLAIMER: DERIVATION!

Well, concentration is a state function, so we can simply choose acid 1 to go first, and acid 2 can go second, suppressed by the equilibrium of the first.

By writing out an ICE table, you would construct the mass action expression for acid 1 (say, $\text{HA}$):

${\text{HA"(aq) rightleftharpoons "H"^(+)(aq) + "A}}^{-} \left(a q\right)$

${K}_{a 1} = \frac{{\left[{\text{H"^(+)]_1["A"^(-)])/(["HA"]) = (["H}}^{+}\right]}_{1} {\alpha}_{1} {C}_{1}}{\left(1 - {\alpha}_{1}\right) {C}_{1}}$

Although we know that $\left[{\text{H"^(+)]_1 = ["A}}^{-}\right]$ in this first process, we choose to write them distinct from each other.

Since we assume ${\alpha}_{1}$ $\text{<<}$ $1$, i.e. ${K}_{a 1} < {10}^{- 5}$ or so, then rewrite this as...

K_(a1) ~~ (["H"^(+)]_1alpha_1C_1)/(C_1) = alpha_1["H"^(+)]_1

So,

${\left[{\text{H}}^{+}\right]}_{1} \approx {K}_{a 1} / {\alpha}_{1}$

The second acid, say ${\text{BH}}^{+}$, now acts.

${\text{BH"^(+)(aq) rightleftharpoons "B"(aq) + "H}}^{+} \left(a q\right)$

And the ${\left[{\text{H}}^{+}\right]}_{1}$ from acid 1 now is the initial concentration for the ${\text{BH}}^{+}$ dissociation.

${K}_{a 2} = \frac{\left[{\text{B"]["H"^(+)])/(["BH"^(+)]) = (["H}}^{+}\right] {\alpha}_{2} {C}_{2}}{\left(1 - {\alpha}_{2}\right) {C}_{2}}$

And since we also have ${\alpha}_{2}$ $\text{<<}$ $1$,

${K}_{a 2} \approx {\alpha}_{2} \left[{\text{H}}^{+}\right]$

$\left[{\text{H}}^{+}\right] \approx {K}_{a 2} / {\alpha}_{2}$

with $\left[{\text{H}}^{+}\right]$ being the net $\left[{\text{H}}^{+}\right]$ concentration. By subtracting out the contribution from acid 1,

${\alpha}_{2} {C}_{2} = {\left[{\text{H}}^{+}\right]}_{2} = {K}_{a 2} / {\alpha}_{2} - {\alpha}_{1} {C}_{1}$.

So, one form of this is

${\left[{\text{H"^(+)] = ["H"^(+)]_1 + ["H}}^{+}\right]}_{2}$

$= {\alpha}_{1} {C}_{1} + {\alpha}_{2} {C}_{2}$

But we have built into ${\alpha}_{2}$ the suppressed equilibrium, and so,

$\textcolor{red}{\left[{\text{H}}^{+}\right] < \frac{{K}_{a 1}}{\alpha} _ 1 + {K}_{a 2} / {\alpha}_{2}}$

It would be convenient to determine ${\alpha}_{2}$ in terms of ${\alpha}_{1}$, but it won't look nice at first.

${K}_{a 2} / {\alpha}_{2} = {\alpha}_{1} {C}_{1} + {\alpha}_{2} {C}_{2}$

$0 = {C}_{2} {\alpha}_{2}^{2} + {\alpha}_{1} {C}_{1} {\alpha}_{2} - {K}_{a 2}$

This becomes a quadratic equation. If you wish to see it,

$\textcolor{g r e e n}{{\alpha}_{2}} = \frac{- \left({\alpha}_{1} {C}_{1}\right) \pm \sqrt{{\alpha}_{1}^{2} {C}_{1}^{2} - 4 {C}_{2} \left(- {K}_{a 2}\right)}}{2 {C}_{2}}$

$= \textcolor{g r e e n}{\frac{- \left({\alpha}_{1} {C}_{1}\right) \pm \sqrt{{\alpha}_{1}^{2} {C}_{1}^{2} + 4 {C}_{2} {K}_{a 2}}}{2 {C}_{2}}}$

And so,

$\left[{\text{H}}^{+}\right] = {\alpha}_{1} {C}_{1} + \left(- \frac{{\alpha}_{1} {C}_{1}}{2 {C}_{2}} \pm \frac{\sqrt{{\alpha}_{1}^{2} {C}_{1}^{2} + 4 {C}_{2} {K}_{a 2}}}{2 {C}_{2}}\right) {C}_{2}$

$= \frac{1}{2} {\alpha}_{1} {C}_{1} \pm \sqrt{{\left(\frac{{\alpha}_{1} {C}_{1}}{2}\right)}^{2} + {K}_{a 2} {C}_{2}}$

And lastly, it would be convenient to know this in terms of ${K}_{a 1}$ instead of ${\alpha}_{1}$, since that requires information we may not already have.

Since ${K}_{a 1} \approx {\alpha}_{1} {\left[{\text{H}}^{+}\right]}_{1} \approx {\alpha}_{1}^{2} {C}_{1}$, we can say that

${\alpha}_{1} = \sqrt{{K}_{a 1} / {C}_{1}}$

Therefore:

$\textcolor{b l u e}{\left[{\text{H}}^{+}\right]} = \frac{1}{2} \sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{{\left(\frac{{C}_{1}}{2}\right)}^{2} {K}_{a 1} / {C}_{1} + {K}_{a 2} {C}_{2}}$

$= \frac{1}{2} \sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{\frac{1}{4} {K}_{a 1} {C}_{1} + {K}_{a 2} {C}_{2}}$

$= \textcolor{b l u e}{\frac{1}{2} \left(\sqrt{{K}_{a 1} {C}_{1}} \pm \sqrt{{K}_{a 1} {C}_{1} + 4 {K}_{a 2} {C}_{2}}\right)}$

TESTING ON ACTUAL PROBLEM

And of course, we should try this on an actual problem.

Consider acetic acid (${K}_{a} = 1.8 \times {10}^{- 5}$) and formic acid (${K}_{a} = 1.8 \times {10}^{- 4}$), for ${C}_{1} = \text{0.50 M}$ and ${C}_{2} = \text{2.00 M}$. Suppose we let formic acid dissociate first.

${\text{HCHO"_2(aq) rightleftharpoons "H"^(+)(aq) + "CHO}}_{2}^{-} \left(a q\right)$

$1.8 \times {10}^{- 4} = {x}^{2} / \left(0.50 - x\right) \approx {x}^{2} / 0.50$

So,

["H"^(+)]_1 ~~ sqrt(1.8 xx 10^(-4) cdot 0.50) = "0.00949 M"

and

${\alpha}_{1} \approx \frac{x}{\left[{\text{HCHO}}_{2}\right]} = 0.019$

Let's see if we get the same ${K}_{a}$.

1.8 xx 10^(-4) stackrel(?)(=) (alpha_1C_1)^2/((1 - alpha_1)C_1)

= (0.019 cdot 0.50)^2/((1 - 0.019)cdot0.50) ~~ 1.8 xx 10^(-4) color(blue)(sqrt"")

${\text{HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O}}_{2}^{-} \left(a q\right)$

1.8 xx 10^(-5) = ((["H"^(+)]_1 + x)(x))/(2.00 - x) ~~ (["H"^(+)]_1 + x)x/2

$\approx {\left[{\text{H}}^{+}\right]}_{1} \frac{x}{2} + {x}^{2} / 2$

The quadratic equation solves to be

x = ["C"_2"H"_3"O"_2^(-)] = "0.00291 M",

which is also the ${\text{H}}^{+}$ contributed by the second acid.

And so, the total ${\text{H}}^{+}$ is:

color(green)(["H"^(+)]) = "0.00949 M" + "0.00291 M" = color(green)ul("0.0124 M")

This acid's percent dissociation at this concentration is

alpha_2 = x/(["HC"_2"H"_3"O"_2]) = "0.00291 M"/"2.00 M" = 0.00145.

So from the first form of the derived equation,

$\textcolor{g r e e n}{\left[{\text{H}}^{+}\right] = {\alpha}_{1} {C}_{1} + {\alpha}_{2} {C}_{2}}$

$= 0.019 \cdot 0.50 + 0.00145 \cdot 2.00 = \textcolor{g r e e n}{\underline{\text{0.0124 M}}}$ color(blue)(sqrt"")

Now to check the general formula we derived above.

color(green)(["H"^(+)]) stackrel(?)(=) 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))

$= \frac{1}{2} \left(\sqrt{1.8 \times {10}^{- 4} \cdot 0.50} \pm \sqrt{1.8 \times {10}^{- 4} \cdot 0.50 + 4 \cdot 1.8 \times {10}^{- 5} \cdot 2.00}\right)$

$= \frac{1}{2} \left(\sqrt{9.00 \times {10}^{- 5}} \pm \sqrt{9.00 \times {10}^{- 5} + 1.44 \times {10}^{- 4}}\right)$

$=$ $\textcolor{g r e e n}{\underline{\text{0.0124 M}}}$ color(blue)(sqrt"")

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