In: Chemistry
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–).
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