Question

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

Concentration of `[H^+]` 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...

Answer 1

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

Note that this works best if `K_(a1)` and `K_(a2)` are both on the order of `10^(-5)` or less, i.e. the acids are sufficiently weak.

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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, `"HA"`):

`"HA"(aq) rightleftharpoons "H"^(+)(aq) + "A"^(-)(aq)`

`K_(a1) = (["H"^(+)]_1["A"^(-)])/(["HA"]) = (["H"^(+)]_1alpha_1C_1)/((1 - alpha_1)C_1)`

Although we know that `["H"^(+)]_1 = ["A"^(-)]` in this first process, we choose to write them distinct from each other.

Since we assume `alpha_1` `"<<"` `1`, i.e. `K_(a1) < 10^(-5)` or so, then rewrite this as...

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

So,

`["H"^(+)]_1 ~~ K_(a1)/alpha_1`

The second acid, say `"BH"^(+)`, now acts.

`"BH"^(+)(aq) rightleftharpoons "B"(aq) + "H"^(+)(aq)`

And the `["H"^(+)]_1` from acid 1 now is the initial concentration for the `"BH"^(+)` dissociation.

`K_(a2) = (["B"]["H"^(+)])/(["BH"^(+)]) = (["H"^(+)]alpha_2C_2)/((1 - alpha_2)C_2)`

And since we also have `alpha_2` `"<<"` `1`,

`K_(a2) ~~ alpha_2["H"^(+)]`

`["H"^(+)] ~~ K_(a2)/alpha_2`

with `["H"^(+)]` being the net `["H"^(+)]` concentration. By subtracting out the contribution from acid 1,

`alpha_2C_2 = ["H"^(+)]_2 = K_(a2)/alpha_2 - alpha_1C_1`.

So, one form of this is

`["H"^(+)] = ["H"^(+)]_1 + ["H"^(+)]_2`

`= alpha_1C_1 + alpha_2C_2`

But we have built into `alpha_2` the suppressed equilibrium, and so,

`color(red)(["H"^(+)] < (K_(a1))/alpha_1 + K_(a2)/alpha_2)`

It would be convenient to determine `alpha_2` in terms of `alpha_1`, but it won't look nice at first.

`K_(a2)/alpha_2 = alpha_1C_1 + alpha_2C_2`

`0 = C_2alpha_2^2 + alpha_1C_1alpha_2 - K_(a2)`

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

`color(green)(alpha_2) = (-(alpha_1C_1) pm sqrt(alpha_1^2C_1^2 - 4C_2(-K_(a2))))/(2C_2)`

`= color(green)((-(alpha_1C_1) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2)))/(2C_2))`

And so,

`["H"^(+)] = alpha_1C_1 + (- (alpha_1C_1)/(2C_2) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2))/(2C_2))C_2`

`= 1/2 alpha_1C_1 pm sqrt(((alpha_1C_1)/(2))^2 + K_(a2)C_2)`

And lastly, it would be convenient to know this in terms of `K_(a1)` instead of `alpha_1`, since that requires information we may not already have.

Since `K_(a1) ~~ alpha_1["H"^(+)]_1 ~~ alpha_1^2C_1`, we can say that

`alpha_1 = sqrt(K_(a1)/C_1)`

Therefore:

`color(blue)(["H"^(+)]) = 1/2 sqrt(K_(a1)C_1) pm sqrt(((C_1)/(2))^2K_(a1)/C_1 + K_(a2)C_2)`

`= 1/2 sqrt(K_(a1)C_1) pm sqrt(1/4K_(a1)C_1 + K_(a2)C_2)`

`= color(blue)(1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2)))`

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TESTING ON ACTUAL PROBLEM

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

Consider acetic acid (`K_a = 1.8 xx 10^(-5)`) and formic acid (`K_a = 1.8 xx 10^(-4)`), for `C_1 = "0.50 M"` and `C_2 = "2.00 M"`. Suppose we let formic acid dissociate first.

`"HCHO"_2(aq) rightleftharpoons "H"^(+)(aq) + "CHO"_2^(-)(aq)`

`1.8 xx 10^(-4) = x^2/(0.50 - x) ~~ x^2/0.50`

So,

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

and

`alpha_1 ~~ x/(["HCHO"_2]) = 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"")`

Next, add in acetic acid.

`"HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O"_2^(-)(aq)`

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

`~~ ["H"^(+)]_1x/2 + x^2/2`

The quadratic equation solves to be

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

which is also the `"H"^(+)` contributed by the second acid.

And so, the total `"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,

`color(green)(["H"^(+)] = alpha_1C_1 + alpha_2C_2)`

`= 0.019 cdot 0.50 + 0.00145 cdot 2.00 = color(green)ul("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))`

`= 1/2 (sqrt(1.8 xx 10^(-4) cdot 0.50) pm sqrt(1.8 xx 10^(-4) cdot 0.50 + 4 cdot 1.8 xx 10^(-5) cdot 2.00))`

`= 1/2(sqrt(9.00 xx 10^(-5)) pm sqrt(9.00 xx 10^(-5) + 1.44 xx 10^(-4)))`

`=` `color(green)ul("0.0124 M")` `color(blue)(sqrt"")`