Question

A 80.0 mL sample of 0.0300 M HClO4 is titrated with 0.0600 M CsOH solution. Calculate the pH after the following volumes of base have been added.

(a) 17.2 mL pH = (b) 39.2 mL pH = (c) 40.0 mL pH =

(d) 40.8 mL pH = (e) 74.4 mL pH =

Answer 1

millimoles of HClO4 = 80 x 0.03 = 2.4

a) millimoles of CsOH added = 17.2 x 0.06 = 1.032

2.4 - 1.032 = 1.368 millimoles HClO4 left

[HClO4] = 1.368 / 97.2 = 0.014 M

pH = -log [H⁺]

pH = -log [0.014]

pH = 1.85

b) millimoles of CsOH added = 39.2 x 0.06 = 2.352

2.4 - 2.352 = 0.048 millimoles HClO4 left

[HClO4] = 0.048 / 119.2 = 0.000403 M

pH = - log [H⁺]

pH = -log [0.000403]

pH = 3.39

c) millimoles of CsOH added = 40 x 0.06 = 2.4

at equivalent point both are strong acid and strong base. so at equivalent point pH = 7

pH = 7.0

d) millimoles of CsOH added = 40.8 x 0.06 = 2.448

2.448 - 2.4 = 0.048 millimoles CsOH left

[CsOH] = 0.0448 / 120.8 = 0.000397 M

pOH = -log [OH^-]

pOH = -log [0.000397]

pOH = 3.40

pH = 14 - 3.40

pH = 10.6

e) millimoles of CsOH added = 74.4 x 0.06 = 4.464

4.464 - 2.4 = 2.064 millimoles CsOH left

[CsOH] = 2.064 / 154.4 = 0.0134 M

pOH = - log [OH^-]

pOH = -log [0.0134]

pOH = 1.87

pH = 14 - 1.87

pH = 12.13