2) In an experiment to determine the molecular weight and the ionization constan

Question

2) In an experiment to determine the molecular weight and the ionization constant for a weak 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. The pH was monitored throughout the titration. The equivalence point was reached when 35.23 milliliters of the base had been added. a) What is true about moles of HA and moles OH added at the equivalence point? b) Given the above information about the equivalence point volume, what is the ½- equivalence point volume? c) What is true about the moles of a weak acid HA and moles of its conjugate base A at the ½-equivalence point of the titration? d) Whatis true about the pH at the ½ 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 ½-equivalence point 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 3) What is the pH of 25.00 mL of 0.100 M ammonia sin 1) before titration begins; 2) after 12.50 mL of 0.100 M HCI is added; 3) after 14.00 mL of 0.100 M HCl is added; 4) after 25.00 mL of the HCl is added; and 5) after 35.00 mi f the HCl is added. pKs of ammonia -4.74.

Explanation / Answer

a)

HA and OH-, in equivalence will be equal

since "equivalence" implies

mmol of HA = mmol of OH-

therefore, it is stoichioemtric ratio HA + OH- = H2O + A-

b)

find half equivalence point

Veq = from

Veq = 35.23 mL

1/2half point --> 35.23/2 = 17.615 mL if the half point

c)

if this is weak acid + storng base titration

then, in the half point

50% of HA has been transofrmed to A-

so we have:

50% of HA left; 50% of A- formed

that is,

mol of HA = mol of A-

d)

in the half point, as stated HA = -A

pH = pKa + log(A-/HA)

will have

pH = pKa + log(1)

pH = pKa

therefore, the half point = pKa

e

molecular weight

MM = mass of acid / mol of acid

MM = 1.3717 g /

mol of HA = mmol = (35.23*0.211) = 7.43353 mmol = 7.43353*10^-3 mol

MM = 1.3717 / (7.43353*10^-3) = 184.52 g/mol

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