Question

# A 30.0 mL sample of 0.05 M HCLO is titrated by 0.0250 M KOH solution. Ka...

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.....

a. the pH when no base has been added

b. the pH when 30.0 mL of the base has been added

c. the pH at the equivalence point

d. the pH when an additional 4.00 mL of the KOH solution has been added beyond the equivalence point

millimoles of HClO = 30 x 0.05 = 1.5

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

a) initially

pH = 1/2[pKa - logC]

pH = 1/2 [7.45 - log 0.05]

pH = 4.37

b) 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 of equivalence point

pH = pKa

pH = 7.45

c) 1.75 millimoles KOH must be added to reach equivalence point

1.75 = V x 0.025

V = 70 mL KOH must be added

total volume = 70 + 30 = 100 mL

[NaClO] = 1.5 / 100 = 0.015 M

pH = 1/2 [pKw + pKa + logC]

pH = 1/2 [14 + 7.45 + log 0.015]

pH = 9.81

d) millimoles of KOH extra = 4 x 0.025 = 0.1

[KOH] = 0.1 / 104 = 0.00096 M

pOH = - log [OH-]

pOH = - log [0.00096]

pOH = 3.02

pH = 14 - 3.02

pH = 10.98

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