In an experiment to determine the molecular weight and the ionization constant f

Question

In an experiment to determine the molecular weight and the ionization constant for a wea acid HA, a student dissolved 1.3717 grams of the acid in water to make 50.00 milliliters of solution. The entire solution was titrated with a 0.2211-molar NaOH solution 1he P ss monitored throughout the titration. The equivalence point was reached when 55.25 miliiter of the base had been added. a) What is true about moles of HA and moles OH added at the equivalence point? b) Given th e above information about the equivalence point volume, what is the h- equivalence point volume? c) What is true about the moles of a weak acid HA and moles of its conjugate base A at the h-equivalence point of the titration? d) What is true about the pH at the h-equivalence point of a titration of a weak acid HA with NaOH? e) From the information above, calculate the molecular weight of the weak acid. f) When 20.00 milliliters of NaOH had been added during the above titration, the pH of the solution was 4.23. How does this 20.00 mL volume compare to the 'h-equivalence point volume from part b)? Based on this, do you expect the pK, to be above or below 4.23? g) Calculate the pK, acid ionization constant for the weak acid from the info in part f) and the initial moles of HA you calculated in part e). h) Calculate the K, for the ascorbate ion, A". i) Calculate the pH of the solution at the equivalence point of the titration.

Explanation / Answer

a) At equivalence point complete neutralization takes place. Hence

moles of HA = moles of OH-

b) According to the given information equivalence point volume is 35.23 ml

Hence half equivalence point volume is 35.23/2 = 17.615 ml

c) At half equivalence point , half of the acid will be neutralised and hence,

moles of weak acid= moles of its conjugate base A-

d) According to Hasselbalch-Henderson equation, pH = pKa + log [A-] / [HA]

At half equivalence point , [A-] = [HA]

Hence pH = pKa

e) At equivalence point moles of HA = moles of OH-

Let moles of acid = m

Hence m x 50 = 0.2211 x 35.23

or, m = 50 / (0.2211 x 35.23) = 0.1558

molarity = moles of solute/ 1 lit

Hence, (50/1000) lit x 0.1558 moles/ lit = 0.00779

Molecular weight of the acid is 1.3717/ 0.00779 = 176.084

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