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How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to...

How many grams of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added to 610.0 mL of a 0.0460 M succinic acid solution to produce a pH of 5.956? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

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Answer #1

Let W g of dipotassium succinate trihydrate (K2C4H4O4·3H2O, MW = 248.32 g/mol) must be added

Number of moles of dipotassium succinate trihydrate

Number of moles of succinic acid (610.0 mL of a 0.0460 M)

0.0281 moles of succinic acid will react with 0.0281 moles of dipotassium succinate trihydrate to form 0.0281 moles of mono potassium succinate.

of dipotassium succinate trihydrate will remain.

Hence, 21.56 g of dipotassium succinate trihydrate must be added.

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