Question

A 0.4352-g sample of an unknown monoprotic acid is dissolved in water and titrated with standardized potassium hydroxide. The equivalence point in the titration is reached after the addition of 31.14 mL of 0.1833 M potassium hydroxide to the sample of the unknown acid. Calculate the molar mass of the acid.

_____ g/mol

Answer 1

moles = concentration (M) X volumell) (1000ml = 1L) KOH = 36.14 mL = 0.0314L moles of KOH = 0.1833M X 0.0314L2 5.708 X10-3 mol let HA is the monoprotic acid. neutralization Renchon HA (aq) + koncaq! acid a base KAlaq) + Heures Imole of KOH neutralize I mole of HA = 5.708 X 10 -3 mole of koH will neutralize 5.708 X10-3 moles of HA = moles of HA present in sample : 5.708 X10-3 mol man of HA lacid) = 0.4352 g molar man of acid - man = 0.4352g. moles 5.708 × 10-3 mol = 76.24 gimol Answer Pleue Rate