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...

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...

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Nov 10, 2017

#["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.


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)))#


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"")#

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