Question

You have prepared a solution by dissolving 0.20 mol glutaric acid (C5H8O2, pKa1 = 4.34, pKa2 = 5.42) and 0.10 mol mandelic acid (C8H8O3, pKa = 3.86) in 1.00 L of water. Write all equilibria occurring in solution as well as mass balance and charge balance expressions for this system. Determine the pH of this solution.

I was suggested to use a spreadsheet. I can work the spreadsheet, but I can't get to equations that describe the whole system.

Answer 1

The pH of a solution containing two acids will depend upon the total concentration of H+ ion relesed from both the acids. Here the given aciids are- glutaric acid (C5H8O2) and mandelic acid (C8H8O3)

Now we know both are weakk acids. So they will partially dissociate to their conjugate base and H3O+ ions

Lets start.

i- We kave glutaric acid (C5H8O2), which is a diprotic acid. IT dissocicates in 2 steps as-

Step 1 :

C5H8O2 + H2O ----------------> H3O+ + C5H7O2- pKa1 = 4.34

So Ka1 = 10-pKa = 10-4.34 = 4.57 * 10-5  

Here Ka1 is the edissociation constant for this step, which is

Ka1 = [H3O+] * [C5H7O2-] / [C5H8O2] = 4.57 * 10-5  

Similalrly

Step 2 :

C5H8O2- + H2O ----------------> H3O+ + C5H6O2-2 pKa2 = 5.42

So Ka2 = 10-pKa2 = 10-5.42 = 3.80 * 10-6  

Here Ka2 is the edissociation constant for this step, which is

Ka2 = [H3O+] * [C5H6O2-2] / [C5H8O2-2] = 3.80 * 10-6  

Now given initial mols of C5H8O2 taken = 0.20 mol

So to find the other concentrations, we have to form the ICE table i.e-

For step 1:  

Reaction	C5H8O2 + H2O ---------------->	H3O+ +	C5H7O2-
Initial	0.20 mol	0	0
Change	-x	+x	+x
Equilibrrium	0.20 - x mol	x	x

Now putting the values for-

Ka1 = [H3O+] * [C5H7O2-] / [C5H8O2]

4.57 * 10-5 =  [x] * [x] / [0.20 - x]

(4.57 * 10-5) * [0.20 - x] = x2  

(0.914 * 10-5) - (4.57 * 10-5)x = x2   

x2 + (4.57 * 10-5)x - (0.914 * 10-5) = 0

Solving this-

x = 0.003

So we can say, here at equilibrium, we have-

H3O+ = x = 0.003 mols

C5H7O2- = x = 0.003 mols

C5H8O2 = 0.20 - x = 0.20 - 0.003 = 0.197‬ mols

Now

For step 2:  

Reaction	C5H7O2- + H2O ---------------->	H3O+ +	C5H7O2-2
Initial	0.003 mol	0.003	0
Change	-x	+x	+x
Equilibrrium	0.003 - x mol	0.003 + x	x

Now putting the values for-

Ka2 = [H3O+] * [C5H7O2-2] / [C5H7O2-]

3.80 * 10-6 =  [0.003 + x] * [x] / [0.003 - x]

(3.80 * 10-6) * [0.003 - x] = [0.003 + x] * [x]

(0.0114‬ * 10-6) - (3.80 * 10-6)x = 0.003x + x2   

x2 + 0.003x + (3.80 * 10-6)x - (0.0114‬ * 10-6) = 0

x2 + 0.0030038x - (0.0114‬ * 10-6) = 0

Solving this-

x = 0.00000379

So we can say, here at equilibrium, we have-

H3O+ = x = 0.003 + x = 0.00000379 + 0.003 = 0.00300379‬ mols

C5H7O2-2 = x = 0.00000379 mols

C5H8O2- = 0.003 - x = 0.003 - 0.00000379 = 0.00299 mols

ii-Now similalry for the 2nd acid-

We kave mandelic acid (C8H8O3), which is a monoprotic acid. IT dissocicates in 1 steps as-

Step 1 :

C8H8O3 + H2O ----------------> H3O+ + C8H7O3- pKa1 = 3.86

So Ka1 = 10-pKa = 10-3.86 = 1.38 * 10-4  

Here Ka1 is the edissociation constant for this step, which is

Ka1 = [H3O+] * [C8H7O3-] / [C8H8O3] = 1.38 * 10-4  

Now given initial mols of C8H8O3 taken = 0.10 mol

So to find the other concentrations, we have to form the ICE table i.e-

For step 1:  

Reaction	C8H8O3 + H2O ---------------->	H3O+ +	C8H7O3-
Initial	0.10 mol	0	0
Change	-x	+x	+x
Equilibrrium	0.10 - x mol	x	x

Now putting the values for-

Ka1 = [H3O+] * [C8H7O3-] / [C8H8O3]

1.38 * 10-4 =  [x] * [x] / [0.10 - x]

(1.38 * 10-4) * [0.10 - x] = x2  

(0.138‬ * 10-4) - (1.38 * 10-4)x = x2   

x2 + (1.38 * 10-4)x - (0.138‬ * 10-4) = 0

Solving this-

x = 0.0036

So we can say, here at equilibrium, we have-

H3O+ = x = 0.0036 mols

C8H7O3- = x = 0.0036 mols

C8H8O3 = 0.10 - x = 0.10 - 0.0036 = 0.0964‬‬ mols

That means now in our solution, total mols of H3O+ presennt = 0.0036 mols + 0.00300379‬ mols

= 0.0066 mols

Again total volume of solution = 1L

So concentration of H3O+ presennt = mols/ volume

= 0.0066 mols/ 1L

= 0.0066 M

So pH = -log [H3O+]

= -log [0.0066]

= 2.18