Question

A 20 mL sample of 0.01 M propionic acid (CH3CH2COOH; Pka = 4.87) is titrated with 0.05 M NaOH.

A) Write out the chemical reaction for this titration. B) Calculate the initial pH of the sample. C) Calculate the volume of NaOH required to reach the equivalence point. D) Calculate the pH of the solution at the equivalence point. E) Sketch a titration curve for this titration (pH versus volume NaOH added). Note the location of the equivalence point on the curve.

Answer 1

a) CH3CH2COOH(aq) + NaOH(aq) ----> CH3CH2COONa(aq) + H2O(l)

b) initial pH = 1/2(pka-logC)

   pka of CH3CH2COOH = 4.87

C = concentration of CH3CH2COOH = 0.01 M

pH = 1/2(4.87-log0.01) = 3.435

c) No of mol of CH3CH2COOH reacted = 20*0.01 = 0.2 mmol

   No of mol of NaOH required = 0.2 mmol

   volume of NaOH required = 0.2/0.05 = 4 ml

d) at equivalence point

concentration of CH3CH2COONa formed = 0.2/(20+4) = 0.00834 M

pH of CH3CH2COONa = 7+1/2(pka+logC)

    pka = 4.87

   C = 0.00834 M

pH of CH3CH2COONa = 7+1/2(4.87+log0.00834) = 8.39

e)  
pH Euivalence point 8.4 . Volume of NaOH added