Question

Calculate the pH for each case in the titration of 50.0 mL of 0.150 M HCIO(aq) with 0.150 M KOH(aq). Use the ionization constant for HCIO. What is the pH before addition of any KOH? pH = What is the pH after addition of 25.0 mL KOH? pH = What is the pH after addition of 30.0 mL KOH? pH = What is the pH after addition of 50.0 mL KOH? pH = What is the pH after addition of 60.0 mL KOH? pH =

Answer 1

ionization constant for HClO is 2.9 * 10-8

pKa = -log (2.9*10^-8) = 7.54

a)

HClO is wea acid.

pH = 0.5 * PKa - 0.5 * log C

pH = 0.5 * - log (2.9 * 10-8) - 0.5 * log (0.150)

pH = 3.77 + 0.412

pH = 4.18

b)

25.0 mL of 0.150 M KOH = 0.025 * 0.150 = 0.00375 mole.

50.0 mL of 0.150 M HClO(aq) = 0.050 * 0.150 = 0.0075 mole.

pH = PKa + log [salt] / [acid] = 7.54 + log (0.00375) / (0.0075 - 0.00375)
pH = 7.54

c)

30.0 mL of 0.150 M KOH = 0.030 * 0.150 = 0.0045 mole.

pH = PKa + log [salt] / [acid] = 7.54 + log (0.00375) / (0.0075 - 0.0045)

pH = 7.64

d)

all the acid is neutralized by base.

so salt hydrolysis occurs.

[salt] = 0.150 / 2 = 0.075 M

pH = 0.5 * PKw + 0.5 * Pka + 0.5 * log C = 7 + 3.77 + 0.5 * log (0.075)

pH = 10.2

e)

excess KOH = 10.0 ml of 0.150 M KOH = 0.010 * 0.150 = 0.0015 mole.

[KOH] = 0.0015 * 1000 / (50 + 60) = 0.0136 M

pOH = - log (0.0136)

pOH = 1.87

pH = 14 - 1.87 = 12.13