The pH of 0.10 M NH3 is closest to11.132.875.138.87

Answers

Answer 1

The pH of 0.10 M NH3 is closest to 11.14.

Option(A)

The Kb value for ammonia, NH3, is 1.8 × 10^-5 at 25°C. The expression for the Kb value is:

Kb = [NH4+][OH-]/[NH3]

where [NH4+] is the concentration of ammonium ion, [OH-] is the concentration of hydroxide ion, and [NH3] is the concentration of ammonia. We can use the relationship Kw = Ka × Kb, where Kw is the ion product constant for water (1.0 × 10^-14 at 25°C) and Ka is the acid dissociation constant for NH4+.

Ka can be calculated from the following equation:

Ka = Kw/Kb = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.56 × 10^-10

At equilibrium, the following reaction occurs:

NH3 + H2O ⇌ NH4+ + OH-

Since NH3 is a weak base, we can assume that its initial concentration is equal to its equilibrium concentration. Therefore, we can use the Kb expression to solve for [OH-]:

Kb = [NH4+][OH-]/[NH3] => [OH-] = Kb[NH3]/[NH4+] = (1.8 × 10^-5)(0.10)/[NH4+]

Since NH4+ is the conjugate acid of NH3, we can assume that it is formed by the reaction of NH3 with water:

NH3 + H2O ⇌ NH4+ + OH-

Therefore, the concentration of NH4+ is equal to [H3O+], and we can use the expression for the acid dissociation constant to solve for [H3O+]:

Ka = [NH4+][H3O+]/[NH3] => [H3O+] = Ka[NH3]/[NH4+] = (5.56 × 10^-10)(0.10)/[NH4+]

Since the solution is in equilibrium, [OH-][H3O+] = Kw = 1.0 × 10^-14. Therefore:

[OH-][H3O+] = (1.8 × 10^-5)(0.10)[NH4+]/[NH3] × (5.56 × 10^-10)(0.10)[NH3]/[NH4+] = 1.0 × 10^-14

Simplifying this equation, we get:

[NH4+][OH-] = 1.0 × 10^-14

Substituting [OH-] = (1.8 × 10^-5)(0.10)/[NH4+] into this equation, we get:

[NH4+] = 0.10/(1.8 × 10^-5)(1.8 × 10^-5)(0.10) = 3.086 M

Finally, we can calculate the pH of the solution using the expression:

pH = -log[H3O+] = -log[(5.56 × 10^-10)(0.10)/3.086] = 11.14    Option(A)

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Answer 2

The pH of 0.10 M NH3 is closest to 11.14. Option(A) Finally, we can calculate the pH of the solution using the expression: pH = -log[H3O+] = -log[(5.56 × 10^-10)(0.10)/3.086] = 11.14    Option(A).

The Kb value for ammonia, NH3, is 1.8 × 10^-5 at 25°C. The expression for the Kb value is:

Kb = [NH4+][OH-]/[NH3]

where [NH4+] is the concentration of ammonium ion, [OH-] is the concentration of hydroxide ion, and [NH3] is the concentration of ammonia. We can use the relationship Kw = Ka × Kb, where Kw is the ion product constant for water (1.0 × 10^-14 at 25°C) and Ka is the acid dissociation constant for NH4+.

Ka can be calculated from the following equation:

Ka = Kw/Kb = (1.0 × 10^-14)/(1.8 × 10^-5) = 5.56 × 10^-10

At equilibrium, the following reaction occurs:

NH3 + H2O ⇌ NH4+ + OH-

Since NH3 is a weak base, we can assume that its initial concentration is equal to its equilibrium concentration. Therefore, we can use the Kb expression to solve for [OH-]:

Kb = [NH4+][OH-]/[NH3] => [OH-] = Kb[NH3]/[NH4+] = (1.8 × 10^-5)(0.10)/[NH4+]

Since NH4+ is the conjugate acid of NH3, we can assume that it is formed by the reaction of NH3 with water:

NH3 + H2O ⇌ NH4+ + OH-

Therefore, the concentration of NH4+ is equal to [H3O+], and we can use the expression for the acid dissociation constant to solve for [H3O+]:

Ka = [NH4+][H3O+]/[NH3] => [H3O+] = Ka[NH3]/[NH4+] = (5.56 × 10^-10)(0.10)/[NH4+]

Since the solution is in equilibrium, [OH-][H3O+] = Kw = 1.0 × 10^-14. Therefore:

[OH-][H3O+] = (1.8 × 10^-5)(0.10)[NH4+]/[NH3] × (5.56 × 10^-10)(0.10)[NH3]/[NH4+] = 1.0 × 10^-14

Simplifying this equation, we get:

[NH4+][OH-] = 1.0 × 10^-14

Substituting [OH-] = (1.8 × 10^-5)(0.10)/[NH4+] into this equation, we get:

[NH4+] = 0.10/(1.8 × 10^-5)(1.8 × 10^-5)(0.10) = 3.086 M

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Not My answer But I saw That You needed it.

Credit zeeshanfazil

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Answers

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