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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 800.0 mL of a 0.0524 M succinic acid solution to produce a pH of 5.969? Succinic acid has pKa values of 4.207 (pKa1) and 5.636 (pKa2).

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Expert Solution

no of moles of succinic acid = molarity * volume in L

                                             = 0.0524*0.8   = 0.04192moles

   PH = Pka2 + log[K2C4H4O4·3H2O]/[H2C4H4O4·3H2O]

5.969 = 4.207 + log[K2C4H4O4·3H2O]/[H2C4H4O4·3H2O]

log[K2C4H4O4·3H2O]/[H2C4H4O4·3H2O] = 5.969-4.207

log[K2C4H4O4·3H2O]/[H2C4H4O4·3H2O]   = 1.762

[K2C4H4O4·3H2O]/[H2C4H4O4·3H2O]         = 101.762

[K2C4H4O4·3H2O]/[H2C4H4O4·3H2O]         = 57.8

no of moles of K2C4H4O4·3H2O                 = 57.8*no of moles H2C4H4O4·3H2O

                                                                    = 57.8*0.04192   = 2.43moles

mass of K2C4H4O4·3H2O        = no of moles * gram molar mass

                                               = 2.43*248.32   = 603.4 g >>>> answer

queen_honey_blossom answered 2 years ago
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