The next few questions will walk you through solving the following problem: At a given temperature, a 5.0M solution of hydrazine (N2H4) as a pH of 11.34. Hydrazine is base. A. What is the concentration of hydroxide ion at equilibrium?
B. What is the pK for hydrazine reacting with water at this temperature?

A. [OH⁻] = 2.188x10⁻³M

B. pKb = 6.02

Explanation:

When hydrazine is in equilbrium with water, its reaction is:

N₂H₄(aq) + H₂O(l) ⇄ HN₂H₄⁺(aq) + OH⁻(aq)

Where Kb, is defined as the ratio between concentrations in equilibrium of the species, thus:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

A. From pH, you can find [OH⁻], thus:

pH = -log [H⁺]

11.34 = -log [H⁺]

4.57x10⁻¹² = [H⁺]

As 1x10⁻¹⁴ = [OH⁻] [H⁺]

1x10⁻¹⁴ / 4.57x10⁻¹² = [OH⁻]

B. Concentrations in equilibrium of the species are:

[N₂H₄] = 5.0M - X

[HN₂H₄⁺] = X

[OH⁻] = X

Where X is reaction coordinate

As [OH⁻] = 2.188x10⁻³M

X = 2.188x10⁻³M

Replacing:

[N₂H₄] = 5.0M - 2.188x10⁻³M = 4.9978M

[HN₂H₄⁺] = 2.188x10⁻³M

[OH⁻] = 2.188x10⁻³M

Replacing in Kb expression:

Kb = [HN₂H₄⁺] [OH⁻] / [N₂H₄]

Kb = [2.188x10⁻³M] [2.188x10⁻³M] / [4.9978M]

Kb = 9.577x10⁻⁷

pKb is defined as -log Kb

pKb = -log 9.577x10⁻⁷

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

i believe the answer is 5,000