Question

A 30.0 mL sample of 0.05 M HCLO is titrated by 0.0250 M KOH solution. Ka for HCLO is 3.5 x 10-8. Calculate

the pH when no base has been added, the pH when 30.0 mL of the base has been added ,. the pH at the equivalence point and the pH when an additional 4.00 mL of the KOH solution has been added beyond the equivalence point

Answer 1

millimoles of HClO = 30 x 0.05 = 1.5

pKa = - log Ka

pKa = - log [3.5 x 10^-8] = 7.45

1) initially

pH = 1/2[pKa - log C]

pH = 1/2 [7.45 - log 0.05]

pH = 4.37

2) millimoles of KOH added = 30 x 0.025 = 0.75

1.5 - 0.75 = 0.75 means half way of equivalence point.

at half way equivalence point

pH = pKa

pH = 7.45

3) at equivalence point all acid convert to salt

1.5 millmoles KOH must be added

1.5 = V x 0.025

V = 60 mL KOH must be added

total volume = 30 + 60 = 90

[salt] = 1.5 / 90 = 0.017 M

pH = 1/2 [pKw + pKa + log C]

pH = 1/2 [14 + 7.45 + log 0.017]

pH = 9.84

4) additional millimoles of KOH = 4.0 x 0.025 = 0.1

total volume = 90 + 4 = 94 mL

[KOH] = 0.1 / 94 = 0.00106 M

as KOH is strong base

pOH = - log [OH^-]

pOH = - log [0.00106]

pOH = 2.97

pH = 14 - 2.97

pH = 11.03