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

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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...

Concentration of

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

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

Note that this works best if

**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"(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

Since we assume

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

So,

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

The second acid, say

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

And the

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

And since we also have

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

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

with

#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 ** suppressed** equilibrium, and so,

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

It would be convenient to determine

#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

Since

#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)))#

**TESTING ON ACTUAL PROBLEM**

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

Consider acetic acid (

#"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

#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**

#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"")#

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