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

when—without warning—something provokes your anger, you may struggle not to succumb to it. since what typically makes you mad is feeling powerless in the face of what seems unfair, your anger is mostly an attempt at a “quick fix” to right the balance. it’s as though you’re raising a fist in protest, proclaiming that you’re not going to capitulate to such injustice.

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be directed toward a person who has intentionally (and needlessly) behaved in a hurtful way toward you; and

be beneficial or advantageous to you (i.e., assist you in achieving a desired goal).

i think you’ll agree that only rarely can you claim that your anger is both warranted and , whether to yourself or the relationship.

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explanation:

first one is the correct answer