Question

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 730.0 mL of a 0.0574 M succinic acid solution to produce a pH of 5.865? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

Answer 1

1) pH of acidic buffer = pka2 + log(K2C4H4O4)/(H2C4H4O4)

no of mol of succinic acid = 0.73*0.0574 = 0.0419 mol

      5.865 = 5.636 + log(x/0.0419)

    x = 0.071

Amount of dipotassium succinate trihydrate = n*Mwt

                        = 0.071*248.32

                        = 17.63 g