Please show me the process of the solution to this problem, thank you!! 4. (2 po

Question

Please show me the process of the solution to this problem, thank you!!

4. (2 points) 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) (0.5 pts) Calculate the molar mass of HA b) (0.5 pts) Is the pH at the equivalence point greater than, less than, your answer. or equal to 7.00. Briefly justify (select one,) c) (1.0 pt) 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 K, of the acid HA?

Explanation / Answer

a)

Molar Mass = mass of acid / moles of acid

mol of acid = mol of base = MV = (0.25)(40*10^-3) = 0.01 mol of base/acid present

then

MM = 1.73/0.01 = 173 g/mol

b)

since this is a WEAK aciid, then, expec thydrolysis

HA + OH- --> H2O + A-

in equilbirium

A- +´H2O <-> HA + OH -

therefore pH > 7, i.e. basic

c)

pH = pKa + log(A-/HA)

initially

mmol of acid = 10

mmol of base = MV = 10*0.25 = 2.5

then, after reactin

mmol of acid = 10-2.5 = 7.5

mmol of base = 2.5

3.51 = pKa + log(2.5/7.5)

pKa = 3.51 -log(2.5/7.5)

pKa = 3.987

Ka = 10^-pKa = 10^-3.987

Ka = 0.00010303

Ka = 1.03*10^-4

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