Answer:
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)
<u>pH = 3.93</u>
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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
<u>pH = 7.52</u>
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)
<u>pH = 7.88</u>
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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
<u>pH = 14 - 3.87 = 10.13</u>
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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
<u>pH = 14 -2 = 12</u>
