Question

10. A titration is performed by adding 0.13 M KOH to 60 mL of 0.363 M HNO₃.

a) Calculate the pH before addition of any KOH.

b) Calculate the pH after the addition of 33.51, 83.77 and 166.54 mL of the base.(Show your work in detail for one of the volumes.)

c) Calculate the volume of base needed to reach the equivalence point.

d) Calculate the pH after adding 5.00 mL of KOH past the equivalence point.

Answer 1

a ) pH before addition of any KOH

  HNO3 molarity = 0.363 M

[H+] = 0.363 M

pH = -log [H+] = -log (0.363)

pH = 0.44

b) after the addition of 33.51 mL of KOH

millimoles of HNO3 = 60 x 0.363 = 21.78

millimoles of KOH = 33.51 x 0.13 = 4.356

acid millimoles > base millimoles . so the solution result is acidic in nature

remaining acid = millimoles difference / total volume

= ( 21.78 -4.356 ) / (60 +33.51)

= 0.186 M

[H+] =  0.186 M

pH = -log [H+]

pH = 0.73

note : similarly calculate for 83.77 and 166.54 mL of the base

c)

at equivalence point millimoles of acid = millimoles of base

21.78 = 0.13 x V

V = 167.5 mL

at equivalence point volume of base = 167.5 mL

d) after adding 5.00 mL of KOH past the equivalence point.

volume of KOH = 167.5 + 5 = 172.5 mL

millimoles of base = 172.5 x 0.13 = 22.43

base millimoles > acid millimoles

[OH-] = ( 22.43 - 21.78 ) / (172.5 + 60)

= 2.80 x 10^-3 M

pOH = -log [OH-]

pOH = -log (2.80 x 10^-3 )

pOH = 2.55

pH + pOH = 14

pH =11.45