Posted: Jan 29, 2016 10:43 pm
So you've got three equilibria, with three pKa values, but instead of giving you the pKa values, they've given you the relative concentration of base over acid in each case. The diprotonated form is the one with the lowest pKa of 1.8 - let's call this H3A2+. The next one is H2A+. The neutral one is HA, and the final one is A-
Ok so far?
So the first [ b ]/[ a ] is: [H2A+] / [H3A2+] = 1 (since the pH = pKa, right?). Let's suppose the relative [H3A2+] is 1.
So [H2A+] = [H3A2+] x 1 = 1
The next [ b ]/[ a ] is given by: [HA] / [H2A+] = 6.31x10-5. Multiplying by [H2A+] gives:
[HA] = [H2A+] 6.31x10-5 = [H3A2+] x 1 x 6.31x10-5
The final [ b ]/[ a ] is given by: [A-] / [HA] = 3.16x10-8. Multiplying by [HA]:
[A-] = [HA] x 3.16x10-8
But as before, we have [HA] from the previous step too, so:
[A-] = [H3A2+] x 1 x 6.31x10-5 x 3.16x10-8
So the relative concentrations of all four species are:
1 : 1 : 6.31x10-5 : 2x10-12
Does that help?
Ok so far?
So the first [ b ]/[ a ] is: [H2A+] / [H3A2+] = 1 (since the pH = pKa, right?). Let's suppose the relative [H3A2+] is 1.
So [H2A+] = [H3A2+] x 1 = 1
The next [ b ]/[ a ] is given by: [HA] / [H2A+] = 6.31x10-5. Multiplying by [H2A+] gives:
[HA] = [H2A+] 6.31x10-5 = [H3A2+] x 1 x 6.31x10-5
The final [ b ]/[ a ] is given by: [A-] / [HA] = 3.16x10-8. Multiplying by [HA]:
[A-] = [HA] x 3.16x10-8
But as before, we have [HA] from the previous step too, so:
[A-] = [H3A2+] x 1 x 6.31x10-5 x 3.16x10-8
So the relative concentrations of all four species are:
1 : 1 : 6.31x10-5 : 2x10-12
Does that help?