A 1.241-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of s

Question

A 1.241-g sample of a solid, weak, monoprotic acid is used to make 100.0 mL of solution. 50.0 mL of this solution was titrated with 0.08126-M NaOH. The pH after the addition of 30.29 mL of base was 4.86, and the equivalence point was reached with the addition of 46.10 mL of base. a) How many millimoles of acid are in the original solid sample? Hint: Don't forget the dilution. mmol acid b) What is the molar mass of the acid? g/mol c) What is the pKa of the acid? pKa =

Explanation / Answer

moles NaOH at the equivalence point = 0.03527 L x 0.07057 M = = 0.002489 moles unknown acid in 60.0 mL= 0.02489 moles unknown acid in 100 mL = 0.02489 x 100 / 60=0.04148 = 41.48 mmoles molar mass unknown acid = 0.926 g/0.04148 mol= 22.3g/mol moles NaOH ( after the addition of 17.00 mL )= 0.0170 L x 0.07057 M=0.00120 moles acid = 0.926 g/ 22.3 g/mol=0.0415 ( in 100 mL) moles acid in 60 mL = 0.0415 x 60/100=0.0249 HA + OH- => A- + H2O moles acid = 0.0249 - 0.00120=0.0237 moles A- = 0.00120 total volume = 60 + 17 = 77 mL = 0.077 L [HA]= 0.0237/ 0.077=0.308 M [A-]= 0.00120 / 0.077 =0.0156 M 7.03 = pKa + log 0.0156/ 0.308 = pKa + 0.0506 pKa = 6.98

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