Determine the pH at the following points in the titration.

Question

A 0.0405 L volume of 0.11 M hypochlorous acid (HClO), a weak acid with K_a = 4.0 ✕ 10⁻⁸, is titrated with 0.11 M rubidium hydroxide (RbOH), a strong base.

(a) before any RbOH has been added (c) after 0.0405 L RbOH has been added

Answer 1

√We will determine the pH at the specified points in the titration of 0.0405 L of 0.11 M hypochlorous acid (HClO) with 0.11 M rubidium hydroxide (RbOH).

Step 1: Initial pH (Before any RbOH is added)

At this point, we only have HClO, which is a weak acid. The pH can be found using the equilibrium expression for weak acids:

[HClO] —> [H⁺] + [ClO^-]

Find the initial concentration of HClO:

[HClO] = 0.11M

Using the Ka expression:

K_a = [H⁺][ClO^-] ⁄ [HClO]

Let x be the concentration of [H⁺] at equilibrium:

4.0 x 10^-8 = x² ⁄ 0.11

Solving for x:

x = √{(4.0 x 10^-8)(0.11)}

x = √(4.4x10^-9)

x = 6.63 x 10^-5M

pH = -log(6.63x10^-5)

pH = 4.18

Step 2: pH after 0.0405 L of RbOH has been added

At this point, we have added an equivalent amount of RbOH, meaning we have reached the equivalence point. The reaction:

HClO + OH^- —> ClO^- + H2O

results in the complete neutralization of HClO, leaving only ClO⁻, which is a weak base.

Find the concentration of ClO⁻ at equivalence point:

Since equal moles of HClO and OH⁻ have reacted, all the acid is converted to its conjugate base ClO⁻.

Initial moles of HClO = 0.0405 x 0.11 = 4.455 x10^-3 moles

Since all of this is now ClO⁻, and the total volume is:

0.0405 + 0.0405 = 0.0810 L

[ClO^-] = 4.455x10^-3moles ⁄ 0.0810L = 0.055 M

Now, we calculate the pH using the hydrolysis of ClO⁻:

K_b = K_w ⁄ K_a = (1.0x10^-14) ⁄ (4.0x10^-8) = 2.5x10^-7

Using the base hydrolysis equation:

Reaction : ClO^- + H₂O —> HClO + OH^-

Initial : 0.055M 0 0

Change : -x +x +x

Equilibrium: 0.055M-x x x

0.055M- x x is here neglected since 0.055 / 2.5x10^-7 is >>>>20

Kb = x*x / 0.055M

2.5x 10^-7 = x²*0.055

Solving for x:

x = √{(2.5x10^-7) (0.055)}

x = √{1.375 x10^-8}

x = 1.17x10^-4M (OH-)

pOH = -log(1.17x10^-4)

pOH = 3.93

pH = 14 - 3.93 = 10.07

Final Answers:

(a) Before any RbOH is added: pH = 4.18

(c) After 0.0405 L of RbOH is added (Equivalence Point): pH = 10.07

Answer 2

(a) To find the pH of 0.11 M HClO prior to the addition of any RbOH, we look at the dissociation/ionization of the weak acid, and apply the Ka equation as follows:

HClO ==> H⁺ + ClO^-

Ka = 4.0x10^-8 = [H⁺][ClO^-] / [HClO]

4.0x10^-8 = (x)(x) / 0.11 - x and since the Ka is so small, we assume x is small also, and rewrite as follows:

4.0x10^-8 = x² / 0.11

x² = 4.4x10^-9

x = [H⁺] = 6.63x10^-5 M (which is only ~ 0.06% of 0.11 so above assumption was valid).

The pH is simply -log [H+] = -log 6.63x10^-5

pH = 4.18

(c) To find the pH after the addition of 0.0405 L of 0.11 M RbOH, we look at the reaction between HClO and RbOH and apply stoichiometry to determine the concentrations of the resulting species.

HClO + RbOH ==> H₂O + RbClO

moles HClO = 0.0405 L x 0.11 mol / L = 4.455x10^-3 mols

moles RbOH = 0.0405 L x 0.11 mol / L = 4.455x10^-3 mols

After addition of RbOH ALL of the HClO has reacting with ALL of the RbOH producing 4.455x10^-3 mols moles of RbClO. The final volume will be 0.0405 L + 0.0405 L = 0.0810 L. Thus...

Final [RbClO] = 4.455x10^-3 mols / 0.0810 L = 0.055 M. This will be the [Rb⁺] and the [ClO^-]. To find the pH of this solution, we look at the hydrolysis of 0.055 M ClO^- as follows:

ClO^- + H₂O ==> HClO + OH^- and we use the Kb for ClO^- which we calculate from the Ka for HClO

Ka x Kb = Kw = 1x10^-14

Kb for ClO^- = 1x10^-14 / 4.0x10^-8 = 2.50x10^-7

2.50x10^-7 = [HClO][OH^-] / [ClO^-] = (x)(x) / 0.055

x² = 1.375x10^-8

x = [OH^-] = 1.17x10^-4

pOH = -log [OH^-] = -log 1.17x10^-4

pOH = 3.93

pH = 14 - 3.93

pH = 10.07