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 780.0 mL of a 0.0348 M succinic acid solution to produce a pH of 5.921? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

Succinic acid (H2C4H4O4) is a diprotic acid and the succinate ion (C2H4O42–) is the completely deprotonated form of succinic acid. The addition of the succinate ion to the solution results in a buffer. Since the desired pH of the solution is greater than pKa2 for succinic acid, we know that the solution contains a mixture of the succinate ion and the monoprotonated intermediate species (HC2H4O4–).

Explanation / Answer

pH = 5.921

pKa2 = 5.636

pH = pKa2 + log [A-] /[HA]

5.921 = 5.636 + log [A-] /[HA]

0.285 = log [A-] /[HA]

[A-] /[HA] = 1.928

[A-] = 1.928 [HA] ------------------------------(1)

moles of total solution = 780.0 x 0.0348 / 1000 = 0.02714

[A-] + [HA] = 0.02714 ---------------------------(2)

from (1) and (2)

1.928 [HA] + [HA]   = 0.02714

2.928 [HA] = 0.02714

[HA] = 0.02714 / 2.928

            = 0.00927

[A-] = 1.928 [HA]

         = 1.928 x 0.00927

         = 0.01787

here   A- is a dipotassium succinate

and HA is monopotassium succinate

[A-] moles = 0.01787

dipotassium succinate trihydrate moles = 0.01787

moles = mass / molar mass

0.01787 = mass / 248.32

mass = 0.02493 x 248.32

          = 4.438 g

mass of dipotassium succinate trihydrate = 4.438 g

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