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

Question

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

Concentration of $[H^{+}]$ $[H^+]$ ion in a mixture of two weak acids... if $α_{1}$ $alpha_1$ and $α_{2}$ $alpha_2$ are their degree of dissociation which are negligible w.r.t unity... $K_{a_{1}}$ $K_(a_1)$ and $K_{a_{2}}$ $K_(a_2)$ are ionization constant of the acids respectively and $C_{1}$ $C_1$ and $C_{2}$ $C_2$ is their concentration...

1 Answer

Impact of this question

1853 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-10T21:13:58

$[H^{+}] = \frac{1}{2} (\sqrt{K_{a 1} C_{1}} \pm \sqrt{K_{a 1} C_{1} + 4 K_{a 2} C_{2}})$ $["H"^(+)] = 1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2))$

Note that this works best if $K_{a 1}$ $K_(a1)$ and $K_{a 2}$ $K_(a2)$ are both on the order of $10^{- 5}$ $10^(-5)$ 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, $HA$ $"HA"$ ):

$HA (a q) ⇌ H^{+} (a q) + A^{-} (a q)$ $"HA"(aq) rightleftharpoons "H"^(+)(aq) + "A"^(-)(aq)$

$K_{a 1} = \frac{{[H^{+}]}_{1}^{+} [A^{-}]}{[HA]} = \frac{{[H^{+}]}_{1}^{+} α_{1} C_{1}}{(1 - α_{1}) C_{1}}$ $K_(a1) = (["H"^(+)]_1["A"^(-)])/(["HA"]) = (["H"^(+)]_1alpha_1C_1)/((1 - alpha_1)C_1)$

Although we know that ${[H^{+}]}_{1}^{+} = [A^{-}]$ $["H"^(+)]_1 = ["A"^(-)]$ in this first process, we choose to write them distinct from each other.

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

$K_{a 1} \approx \frac{{[H^{+}]}_{1}^{+} α_{1} C_{1}}{C_{1}} = α_{1} {[H^{+}]}_{1}^{+}$ $K_(a1) ~~ (["H"^(+)]_1alpha_1C_1)/(C_1) = alpha_1["H"^(+)]_1$

So,

${[H^{+}]}_{1}^{+} \approx \frac{K_{a 1}}{α_{1}}$ $["H"^(+)]_1 ~~ K_(a1)/alpha_1$

The second acid, say ${BH}^{+}$ $"BH"^(+)$ , now acts.

${BH}^{+} (a q) ⇌ B (a q) + H^{+} (a q)$ $"BH"^(+)(aq) rightleftharpoons "B"(aq) + "H"^(+)(aq)$

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

$K_{a 2} = \frac{[B] [H^{+}]}{[{BH}^{+}]} = \frac{[H^{+}] α_{2} C_{2}}{(1 - α_{2}) C_{2}}$ $K_(a2) = (["B"]["H"^(+)])/(["BH"^(+)]) = (["H"^(+)]alpha_2C_2)/((1 - alpha_2)C_2)$

And since we also have $α_{2}$ $alpha_2$ $<<$ $"<<"$ $1$ $1$ ,

$K_{a 2} \approx α_{2} [H^{+}]$ $K_(a2) ~~ alpha_2["H"^(+)]$

$[H^{+}] \approx \frac{K_{a 2}}{α_{2}}$ $["H"^(+)] ~~ K_(a2)/alpha_2$

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

$α_{2} C_{2} = {[H^{+}]}_{2}^{+} = \frac{K_{a 2}}{α_{2}} - α_{1} C_{1}$ $alpha_2C_2 = ["H"^(+)]_2 = K_(a2)/alpha_2 - alpha_1C_1$ .

So, one form of this is

$[H^{+}] = {[H^{+}]}_{1}^{+} + {[H^{+}]}_{2}^{+}$ $["H"^(+)] = ["H"^(+)]_1 + ["H"^(+)]_2$

$= α_{1} C_{1} + α_{2} C_{2}$ $= alpha_1C_1 + alpha_2C_2$

But we have built into $α_{2}$ $alpha_2$ the suppressed equilibrium, and so,

$[H^{+}] < \frac{K_{a 1}}{α_{1}} + \frac{K_{a 2}}{α_{2}}$ $color(red)(["H"^(+)] < (K_(a1))/alpha_1 + K_(a2)/alpha_2)$

It would be convenient to determine $α_{2}$ $alpha_2$ in terms of $α_{1}$ $alpha_1$ , but it won't look nice at first.

$\frac{K_{a 2}}{α_{2}} = α_{1} C_{1} + α_{2} C_{2}$ $K_(a2)/alpha_2 = alpha_1C_1 + alpha_2C_2$

$0 = C_{2} α_{2}^{2} + α_{1} C_{1} α_{2} - K_{a 2}$ $0 = C_2alpha_2^2 + alpha_1C_1alpha_2 - K_(a2)$

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

$α_{2} = \frac{- (α_{1} C_{1}) \pm \sqrt{α_{1}^{2} C_{1}^{2} - 4 C_{2} (- K_{a 2})}}{2 C_{2}}$ $color(green)(alpha_2) = (-(alpha_1C_1) pm sqrt(alpha_1^2C_1^2 - 4C_2(-K_(a2))))/(2C_2)$

$= \frac{- (α_{1} C_{1}) \pm \sqrt{α_{1}^{2} C_{1}^{2} + 4 C_{2} K_{a 2}}}{2 C_{2}}$ $= color(green)((-(alpha_1C_1) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2)))/(2C_2))$

And so,

$[H^{+}] = α_{1} C_{1} + ⎛ ⎜ ⎜ ⎝ - \frac{α_{1} C_{1}}{2 C_{2}} \pm \frac{\sqrt{α_{1}^{2} C_{1}^{2} + 4 C_{2} K_{a 2}}}{2 C_{2}} ⎞ ⎟ ⎟ ⎠ C_{2}$ $["H"^(+)] = alpha_1C_1 + (- (alpha_1C_1)/(2C_2) pm sqrt(alpha_1^2C_1^2 + 4C_2K_(a2))/(2C_2))C_2$

$= \frac{1}{2} α_{1} C_{1} \pm \sqrt{{(\frac{α_{1} C_{1}}{2})}^{2} + K_{a 2} C_{2}}$ $= 1/2 alpha_1C_1 pm sqrt(((alpha_1C_1)/(2))^2 + K_(a2)C_2)$

And lastly, it would be convenient to know this in terms of $K_{a 1}$ $K_(a1)$ instead of $α_{1}$ $alpha_1$ , since that requires information we may not already have.

Since $K_{a 1} \approx α_{1} {[H^{+}]}_{1}^{+} \approx α_{1}^{2} C_{1}$ $K_(a1) ~~ alpha_1["H"^(+)]_1 ~~ alpha_1^2C_1$ , we can say that

$α_{1} = \sqrt{\frac{K_{a 1}}{C_{1}}}$ $alpha_1 = sqrt(K_(a1)/C_1)$

Therefore:

$[H^{+}] = \frac{1}{2} \sqrt{K_{a 1} C_{1}} \pm \sqrt{{(\frac{C_{1}}{2})}^{2} \frac{K_{a 1}}{C_{1}} + K_{a 2} C_{2}}$ $color(blue)(["H"^(+)]) = 1/2 sqrt(K_(a1)C_1) pm sqrt(((C_1)/(2))^2K_(a1)/C_1 + K_(a2)C_2)$

$= \frac{1}{2} \sqrt{K_{a 1} C_{1}} \pm \sqrt{\frac{1}{4} K_{a 1} C_{1} + K_{a 2} C_{2}}$ $= 1/2 sqrt(K_(a1)C_1) pm sqrt(1/4K_(a1)C_1 + K_(a2)C_2)$

$= \frac{1}{2} (\sqrt{K_{a 1} C_{1}} \pm \sqrt{K_{a 1} C_{1} + 4 K_{a 2} C_{2}})$ $= color(blue)(1/2 (sqrt(K_(a1)C_1) pm sqrt(K_(a1)C_1 + 4K_(a2)C_2)))$

TESTING ON ACTUAL PROBLEM

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

Consider acetic acid ( $K_{a} = 1.8 \times 10^{- 5}$ $K_a = 1.8 xx 10^(-5)$ ) and formic acid ( $K_{a} = 1.8 \times 10^{- 4}$ $K_a = 1.8 xx 10^(-4)$ ), for $C_{1} = 0.50 M$ $C_1 = "0.50 M"$ and $C_{2} = 2.00 M$ $C_2 = "2.00 M"$ . Suppose we let formic acid dissociate first.

${HCHO}_{2} (a q) ⇌ H^{+} (a q) + {CHO}_{2}^{-} (a q)$ $"HCHO"_2(aq) rightleftharpoons "H"^(+)(aq) + "CHO"_2^(-)(aq)$

$1.8 \times 10^{- 4} = \frac{x^{2}}{0.50 - x} \approx \frac{x^{2}}{0.50}$ $1.8 xx 10^(-4) = x^2/(0.50 - x) ~~ x^2/0.50$

So,

${[H^{+}]}_{1}^{+} \approx \sqrt{1.8 \times 10^{- 4} \cdot 0.50} = 0.00949 M$ $["H"^(+)]_1 ~~ sqrt(1.8 xx 10^(-4) cdot 0.50) = "0.00949 M"$

and

$α_{1} \approx \frac{x}{[{HCHO}_{2}]} = 0.019$ $alpha_1 ~~ x/(["HCHO"_2]) = 0.019$

Let's see if we get the same $K_{a}$ $K_a$ .

$1.8 \times 10^{- 4} ? = \frac{{(α_{1} C_{1})}_{1}^{2}}{(1 - α_{1}) C_{1}}$ $1.8 xx 10^(-4) stackrel(?)(=) (alpha_1C_1)^2/((1 - alpha_1)C_1)$

$= \frac{{(0.019 \cdot 0.50)}^{2}}{(1 - 0.019) \cdot 0.50} \approx 1.8 \times 10^{- 4} \sqrt{}$ $= (0.019 cdot 0.50)^2/((1 - 0.019)cdot0.50) ~~ 1.8 xx 10^(-4) color(blue)(sqrt"")$

Next, add in acetic acid.

${HC}_{2} H_{3} O_{2} (a q) ⇌ H^{+} (a q) + C_{2} H_{3} O_{2}^{-} (a q)$ $"HC"_2"H"_3"O"_2(aq) rightleftharpoons "H"^(+)(aq) + "C"_2"H"_3"O"_2^(-)(aq)$

$1.8 \times 10^{- 5} = \frac{({[H^{+}]}_{1}^{+} + x) (x)}{2.00 - x} \approx ({[H^{+}]}_{1}^{+} + x) \frac{x}{2}$ $1.8 xx 10^(-5) = ((["H"^(+)]_1 + x)(x))/(2.00 - x) ~~ (["H"^(+)]_1 + x)x/2$

$\approx {[H^{+}]}_{1}^{+} \frac{x}{2} + \frac{x^{2}}{2}$ $~~ ["H"^(+)]_1x/2 + x^2/2$

The quadratic equation solves to be

$x = [C_{2} H_{3} O_{2}^{-}] = 0.00291 M$ $x = ["C"_2"H"_3"O"_2^(-)] = "0.00291 M"$ ,

which is also the $H^{+}$ $"H"^(+)$ contributed by the second acid.

And so, the total $H^{+}$ $"H"^(+)$ 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.

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

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

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

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

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question