Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 760.0 mL of a 0.0583 M succinic acid solution to produce a pH of 5.980? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

Answer 1

dipotassium succinate trihydrate and succinic acid solution act as a buffer solution. Since the pH is greater than pKa2 of the acid, the buffer action is due to interconversion of K2C4H4O4(aq) to KHC4H4O4(aq)

The desired pH = 5.980 => [H3O+] = 1.05x10^-6 M

Given the concentration of succinic acid(H2C4H4O4), [H2C4H4O4] = 0.0583 M

Here KHC4H4O4 is formed both due to dissociation of succinic acid and the hyrolysis of K2C4H4O4.

For the first dissociation of succinic acid

H2C4H4O4 -----> HC₄H₄O₄^- + H3O+, Ka1 = 6.21x10^-5

(0.0583 - X)M ----- X M, -------- X M

Ka1 = 6.21x10^-5 = X² / (0.0583 - X)

=> X = [HC₄H₄O₄^-] = 0.001872 M

HC₄H₄O₄^- lost due to 2nd dissociation is very small and can be neglected.

Let the mass of dipotassium succinate trihydrate, K2C4H4O4·3H2O taken = 'm' g

moles of dipotassium succinate trihydrate, K2C4H4O4·3H2O = m / 248.32 g/mol = m / 248.32 mol

Concentration, [K2C4H4O4·] = m / 248.32 mol / 0.760L = m / 188.723 M

Applying Hendersen equation

pH = pKa2 + log [K2C4H4O4] / [KHC4H4O4]

=> 5.980 = 5.636 + log[K2C4H4O4] / [KHC4H4O4]

=>  log[K2C4H4O4] / [KHC4H4O4] = 0.344

=> [K2C4H4O4] / [KHC4H4O4] = 2.208

=> m / 188.723 M /  0.001872 M = 2.208

=> m = 0.78 g (answer)