Calculate the pH for each case in the titration of 50.0 mL of 0.110 M HClO ( aq ) with 0.110 M KOH ( aq ) . Use the ionization constant for HClO . 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 35.0 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 =

Before KOH is added: pH = 3.93

After adding 25.0 mL KOH: pH = 7.52

After ading 35.0 mL KOH: pH = 7.88

After adding 50.0 mL KOH: pH = 10.13

After adding 60.0 mL KOH: pH = 12

Explanation:

Step 1: Data given

Volume of HClO = 50.0 mL = 0.050 L

Molarity of HClO = 0.110 M

Molarity of KOH = 0.110 M

Ka of HClO = 3.0 * 10^-8

Kb = Kw / Ka = 10^-14 / 3.0 * 10^-8 = 3.33 * 10^-7

Before KOH is added

HClO + H2O → ClO- + H3O+

Calculate the initial concentrations

[HClO] = 0.110 M * 0.050 L = 0.455 M

[ClO-] = 0M

[H3O+] = 0M

The concentration at the equilibrium

[HClO] = 0.455 - XM

[ClO-] = XM

[H3O+] = XM

Ka= [ClO-][H3O+] / [HClO]

3.0 * 10^-8 = X² / 0.455 - X

3.0 * 10^-8 = X² / 0.455

X² = 1.365*10^-8

X=0.000117  = [H3O+]

pH = - log(0.000117)

pH = 3.93

pH after addition of 25.0 mL of KOH

Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

Moles KOH = 0.025 L * 0.11 M = 0.00275 moles

The limiting reactant is KOH. It will completely be consumed (0.00275 moles). HClO is in excess. There will react 0.00275 moles. There will remain 0.00550 - 0.00275 = 0.00275 moles.

Moles KClO (ClO-) produced = 0.00275 moles

pH = pKa + log[ClO-]/[HClO]

pH = 7.52 + 0

pH = 7.52

pH after addition of 35.0 mL of KOH

Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

Moles KOH = 0.035 L * 0.11 M = 0.00385 moles

The limiting reactant is KOH. It will completely be consumed (0.00385 moles). HClO is in excess. There will react 0.00385 moles. There will remain 0.00550 - 0.00385 = 0.00165 moles

Moles KClO (ClO-) produced = 0.00385 moles

[HClO] = 0.00165 moles / 0.085 L =  0.0194 M

[ClO-] = 0.00385 moles / 0.085 L = 0.0453 M

pH = pKa + log[ClO-]/[HClO]

pH = 7.52 + log(0.0453/0.0194)

pH = 7.88

pH after addition of 50.0 mL of KOH

ClO- + H2O ⇄ HClO + OH-

Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

Moles KOH = 0.050 L * 0.11 M = 0.00550 moles

Both reactants will be consumed, there are no limiting reactants. No HClO or KOH will be left.

There will be 0.00550 moles ClO- produced

[ClO-] = 0.00550 moles / 0.100 L = 0.055 M

The initial concentrations:

[ClO-] = 0.055 M

[HClO] = 0 M

[OH-] = 0M

Cocnentration at the equilibrium

[ClO-] = 0.055 - X M

[HClO] = X M

[OH-] = XM

Kb = [HClO][OH-] / [ClO-]

3.33 * 10^-7 = X² / 0.055 - X

3.33 * 10^-7 = X² / 0.055

X² = 1.83*10^-8

X = 0.0001354 = [OH-]

pOH = -log(0.0001354)

pOH = 3.87

pH = 14 - 3.87 = 10.13

pH after addition of 60.0 mL of KOH

Moles HClO = 0.050 L * 0.11 M = 0.00550 moles

Moles KOH = 0.060 L * 0.11 M = 0.00660 moles

We have 0.0011 moles of KOH remaining. KOH is a strong base

[KOH] = 0.0011 moles / 0.11 L

[KOH] = 0.01 M

pOH = -log(0.01)

pOH = 2

pH = 14 -2 = 12