A) One student discovers that 40.60 mL of the base solution is required to titra

Question

A) One student discovers that 40.60 mL of the base solution is required to titrate 0.5005g to KHP to a phenolpthalein end pt. In a diff titration, 26.50 mL of this same base solution is required to titrate 0.4004g of an unknown monoprotic acid. Determine the molarity of the base solution.

How I went about solving the problem: I found the # of moles of base then found the molarity. Did the same thing for the unknown than subtracted both of the found molarities. We were told in lab that KHP is 204.23 g/mol.

B) When calculating the molarity of the NaOH, a student assumes that KHP is a diprotic acid. Would the exp molar mass be too large, small, or unaffected and why?

Explanation / Answer

(A) Let the base be NaOH

NaOH + KHP => NaKP + H2O

Moles of NaOH = moles of KHP = mass/molar mass of KHP

= 0.5005/204.23 = 0.0024507 mol

Molarity of NaOH = moles/volume of base

= 0.0024507/0.04060

= 0.06036 M


Let the acid be HA

NaOH + HA => NaA + H2O

Moles of HA = moles of NaOH = volume x concentration of NaOH

= 26.50/1000 x 0.06036 = 0.0015996 mol


Molar mass of HA = mass/moles of HA

= 0.4004/0.0015996

= 250.3 g/mol


(B) 2 NaOH + KH2P => Na2KP + 2 H2O

NaOH + HA => NaA + H2O

Moles of HA = Moles of NaOH = 2 x moles of KH2P

=> calculated moles of NaOH and HA will be too large (twice the correct value)

=> calculated molar mass of HA = mass/moles of HA will be too small (half the correct value)

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