A 1.73 g sample of a pure monoprotic acid, HA , was dissolved in distilled water
ID: 972916 • Letter: A
Question
A 1.73 g sample of a pure monoprotic acid, HA , was dissolved in distilled water, to a create a solution with volume 50.0 mL. The HA solution was then titrated with 0.250 M NaOH . The pH was measured throughout the titration, and the equivalence point was reached when 40.0 mL of the NaOH solution had been added.
a) Calculate the molar mass of HA .
b) Is the pH at the equivalence point greater than, less than, or equal to 7.00? Briefly justify your answer. (select one)
c) At the 10.0 mL position in the titration (i.e. when 10.0 mL of 0.250 M NaOH have been added to the 50.0 mL of HA solution), the pH is 3.51. What is Ka of the acid HA?
Explanation / Answer
a) mole of acid = moles of base at equivalence point
moles of base = 0.250 x 40 / 1000 = 0.01
moles of acid = 0.01
moles = mass / molar mass
0.01 = 1.73 / molar mass
molar mass = 173 g/mol
molar mass of HA = 173 g/mol
b)
pH is greater than 7 at the equivalence point . because only salt NaA formed here. this salt is from weak acid and strong base. so it is basic salt
c) molarity of HA = moles / volume = 0.01 / (50 x 10^-3) = 0.2 M
C = 0.2 M
[H+] = 10^-pH = 10^-3.51 = 3.1 x 10^-4 = x
Ka = C x^2
Ka = 0.2 x ( 3.1 x 10^-4)^2
Ka = 1.91 x 10^-8
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.