How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g

Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 760.0 mL of a 0.0368 M succinic acid solution to produce a pH of 5.812? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2). Mass:____ g

Hint: Succinic acid (H2C4H4O4) is a diprotic acid that dissociates when placed in water according to the following reactions.

Since the target pH of the solution is greater than pKa2 of succinic acid, we know the solution at this pH will contain a mixture of the succinate ion (C4H4O42–) and the monoprotonated intermediate ion (HC4H4O4–). To achieve this composition sufficient succinate ion must be added to the solution to completely convert all of the succinic acid to the intermediate ion. The reaction between these two species is shown below. Pay attention to the stoichiometry of this reaction.

H2C4H4O4 + C4H4O4^2- --> 2HC4H4O4^-

After this initial addition of the succinate ion, the solution will contain only the intermediate ion. Use the appropriate Henderson-Hasselbalch equation to determine how much additional succinate must be added to the solution to achieve a pH of 5.812. The total number of moles of succinate added to the solution will be the sum of these two steps.

Explanation / Answer

For buffer of succinic acid/sodium succinate

pH = pKa + log(bse/acid)

let x be the moles of sodium succinate added

5.812 = 5.636 + log(x/(0.0368 M x 760 ml - x))

42 - 1.5x = x

x = 42/2.5 = 16.8 mmol

So grams of sodium succinate trihydrate to be added = 16.8 x 248.32/1000 = 4.172 g

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