Question

Dissolve 1 mole of Succinic Acid into 1 Liter of water. What is the pH of the solution? What is the molarity of the protonated and depronated forms of Succinic Acid?

Answer 1

M = 1M of Acid of H2A

pKa1 = 4.16

pKa2 = 5.61

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] =1 - x

Where x is the acid already in solution

Ka1 = 10^-4.16 = 6.92*10^-5

Substitute in Ka1

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

6.92*10^-5= x*x / (1 - x )

Solve for x

x = 0.00828 and -0.0083

ignore the negative solution since there are no negative concentrations

x = [H+] = [HA-] = 0.00828

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 = 10^-5.61 = 2.45*10^-6

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

Substitute all data

2.45*10^-6 = y*y / (x-y)

2.45*10^-6= y² / ( 0.00828-y)

Sovle for y

y = 1.41*10^-4 and -1.43*10^-4

Ignore negative values, no negative concentrations exist

Then

[H+] = y = 1.41*10^-4

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

that is x+y = 0.00828+ 1.41*10^-4 = 0.008421

As you can see, the effect of the second ion is not enough to change the value of x significantively

pH = -log[H+] = -log(0.008421) = 2.07

pH = 2.07

[H+] = x+y = 0.0004768

[A-2] = y = 1.41*10^-4

[HA-] = x-y = 0.00828 - 1.41*10^-4 = 0.008139

[H2A] = 1-x = 1-0.00828 = 0.99172