Answer to Question #118036 in Chemistry for Crizzia Paula Lobrido

Question

Determine the Concentration of H+ , OH-, pH, and pOH of the following solutions:

1. 0.0100 M HCl
2. 0.0200 M HOCl [ HOCl(aq) + H2O(l) ⇄ H3O+(aq) + OCl-(aq); Ka = 3.0 x 10 -8]

Answer 1

Solutiom (1):

When HCl molecules dissolve they dissociate into H⁺ ions and Cl^- ions. HCl is a strong acid because it dissociates almost completely.

HCl(aq) = H⁺(aq) + Cl^-(aq)

Therefore,

C(HCl) = [H⁺] = 0.0100 M

[H⁺] = 0.01 M

We can convert between [H⁺] and pH using the following equations:

pH = - log[H⁺]

pH = - log(0.0100) = 2

pH = 2

For any aqueous solution at 25^∘C:

pH + pOH = 14

pOH = 14 - pH = 14 - 2 = 12

pOH = 12

We can convert between [OH^-] and pOH using the following equations:

pOH = - log[OH^-]

[OH^-] = 10^-pOH

[OH^-] = 10^-12 = 1×10^-12 M

[OH^-] = 1×10^-12 M

Answer (1): [H⁺] = 0.01 M; pH = 2; pOH = 12; [OH^-] = 1×10^-12 M.

Solution (2):

Since hypochlorous acid (HOCl) is a weak acid and remains mostly undissociated, the major species in a 0.0200 M HOCl solution are HOCl and H₂O.

Both species can produce H⁺. Since HOCl is a significantly stronger acid than H₂O, we can neglect water as a source of H⁺.

The dominant equilibrium is:

HOCl(aq) + H₂O(l) ⇄ H₃O⁺(aq) + OCl^-(aq)

or

HOCl(aq) = H⁺(aq) + OCl^-(aq)

We therefore use the following equilibrium expression:

The concentrations are as follows:

[HOCl] = 0.02 - x

[H⁺] = [OCl^-] = x

Then,

Assuming that 0.02 >> x, then:

(0.02 - x) = 0.02

K_a = x² / 0.02

x² = (3.0×10^-8 × 0.02) = 6×10^-10

x = (K_a)^0.5 = (6×10^-10)^0.5 = 2.45×10^-5

x= [H⁺] = 2.45×10^-5 M

[H⁺] = 2.45×10^-5 M

We can convert between [H⁺] and pH using the following equations:

pH = - log[H⁺]

pH = - log(2.45×10^-5) = 4.61

pH = 4.61

For any aqueous solution at 25^∘C:

pH + pOH = 14

pOH= 14 - pH = 14 - 4.61 = 9.39

pOH = 9.39

We can convert between [OH-] and pOH using the following equations:

pOH = - log[OH^-]

[OH^-] = 10^-pOH

[OH^-] =10^-9.39 = 4.07×10^-10 M

[OH^-] = 4.07×10^-10 M

Answer (2): [H⁺] = 2.45×10^-5 M; pH = 4.61; pOH = 9.39; [OH^-] = 4.07×10^-10 M.