How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g
ID: 939711 • Letter: H
Question
How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 640.0 mL of a 0.0568 M succinic acid solution to produce a pH of 5.960? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).Mass= How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 640.0 mL of a 0.0568 M succinic acid solution to produce a pH of 5.960? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).
Mass= How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 640.0 mL of a 0.0568 M succinic acid solution to produce a pH of 5.960? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).
Mass=
Explanation / Answer
No of mol of succinic acid = 0.0568*0.64 = 0.036 mol
pH = pka + log(salt/acid)
net pka of succinic acid = (4.207+5.536) / 2 = 4.8715
5.96 = 4.8715+log(x/0.036)
x = 0.44 mol
No of mol of succinate trihydrate required = 0.44 mol
mass of succinate trihydrate required = 0.44*248.32 = 109.26 grams
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