Question

Calculate the concentrations of all species contained in 0.15 M solution of carbonic acid at 25 degrees celsius. ka1= 4.2 x10^-7, ka2= 4.8 x 10^-11 .

Answer 1

Let H2A = H2CO3 and HA- = HCO3- and A-2 = CO3-2

M = 0.15 M of H2A

H2A <-> H+ HA-

HA- <-> H+ A^-2

This will be formed...

We need the [H+] concentration

From the first equilbrium

Ka1 = [HA-][H+] /[H2A]

Ka2 = [A^-2 ][H+]/ [HA-]

There will be two H+ to account, the Ka1 and form Ka2

Let us calculate first the ammount of Ka1 H+ ions

NOTE: [HA-] = [H+] = x since for 1 mol of [H+] you will have always another mol of HA-

[H2A] = 0.15 - x

Where x is the acid already in solution

Ka1 = 4.2*10^-7

Substitute in Ka1

Ka1 = [HA-][H+] /[H2A]

4.2*10^-7 = x*x / (0.15 - x )

Solve for x

x = 2.5*10^-4 and -2.51*10^-4

ignore the negative solution since there are no negative concentrations

x = [H+] = [HA-] = 2.5*10^-4

Please remember this number, we will use it at the end when calculating total H+ moles

HINT: we could ignore the second equilibrium since Ka2 is too low compared to the Ka1. That is, the 95% or more of the acid is given by the first ionization

But lets do the other Ka2 equilibrium

Assume, once again

[A^-2 ]= [H+] = y NOTE that these H+ ions are from the SECOND ionization and cannot be compared with the first ionization, thats why I'm using x and y to denote difference

[HA-] = x-y

[HA-] = x will be before the second dissociation, after the dissociation you need to account for the lost acid which is denoted as -y

Ka2 = 4.8*10^-11

Ka2 = [A^-2 ][H+]/ [HA-]

Substitute all data

4.8*10^-11 = y*y / (x-y)

4.8*10^-11= y² / ( 2.5*10^-4-y)

Solve for y

y = 1.09*10^-7 and -1.09*10^-7

Ignore negative values, no negative concentrations exist

Then

[H+] = y = 1.09*10^-7

Time to add [H+] of first ionization +[H+] of second ionization

that is x+y = 2.5*10^-4+ 1.09*10^-7 = 2.50*10^-4

[H+] = x+y = 2.50*10^-4

[A-2] = [CO3^-2] =y = 1.09*10^-7

[HA-] = [HCO3^-] = x-y = 2.5*10^-4-1.09*10^-7 = 2.4989*10^-4

[H2A] = [H2CO3] = 0.15-x = 0.15-2.5*10^-4 = 0.14975

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Monoprotic acid

HB is a weak acid, it will form an equilibrium when in solution:

HB <-> H+ and B-

Note that there is 1 mol of H+ for every mol of B-

Since Ka = 10^-pka = 10^-4.2

Ka = 6.31*10^-5

The equilibrium expression is given by:

Ka = [H+][B-] / [HB]

Let us assume:  [H+]= [B-] = x ... since 1 mol of H+ is present for every mol of B-, the concentrations must be the same

Not only that. Since we are ionizing some acid, we must correct the concentration of [HB] as follows:

[HB] = 0.77 -x where the "-x" corrects the dissociated acid. Since it is a weak acid, in general, you may ignore this. For this proble I will not ignore it but you may do it to simplify math

Substitute all in Ka expression

Ka = [H+][B-] / [HB]

6.31*10^-5 = x*x/(0.77-x)

Solve for x

x = 0.0069 and -0.0070.... ignore negative value since there are no negative concentrations

and since x = [H+] = 0.0069

pH = -log[H+] = -log(0.0069) = 2.16

And the % of ionization is = ionied / total *100% = 0.0069 / 0.77 *100% = 0.8961 % which is a very small amount