Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 690.0 mL of a 0.0526 M succinic acid solution to produce a pH of 5.818? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

____?____grams

Answer 1

The deprotonation reactions of succinic acid are

with (1)

with (2)

So the dissociation constants for succinic acid are K_a1 = 6.209 x 10^-5 and K_a2 = 2.312 x 10^-6.

Since the dissociation constants are so close, we cannot ignore any one of them but must treat them together unlike other diprotic acids where one dissociation constant is orders of magnitude larger than the other constant.

The moles of succinic acid in 690 mL is given by

   (3)

Let n_salt be the number of moles of the potassium succinate added.

Combining equations (1) and (2) and after some re-arrangement, we get

(4)

So we can make an ICE chart for the above reaction as

I 0.036 mol n_salt mol 0 mol

C -x mol +x mol +2x mol

E 0.036 -x mol n_salt +x mol 2x mol

Here, it is assumed that initially all the succinate anions are from the salt added and none from the acid.

For equation (4), the equlibrium constant is given by

(5)

We can use equation (5) to solve for n_salt in terms of 'x' from the ICE chart with the volume of the system being 0.69 L.

Therefore,

   (7)

(8)

Thus, the moles of the species are 2x mol.

We can use the Henderson–Hasselbalch equation to get the concentration of succinate in the second deprotonation step as shown in equation (2) since the pKa for this step is closest to the desired pH.

(9)

Now using equation (8) in equation (9), we can solve for x as follows,

   (10)

Using the value of x from equation (10) in equation (8), we get the number of moles of the succinate added as the potassium salt as

   (11)

Thus the mass of the potassium succinate added will be

(8)