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NH3 is a weak base ( b=1.8×10−5 ) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.036 M in NH4Cl at 25 °C?

User Cornstalks
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Answer: The pH of a 0.036 M solution of NH4Cl at 25°C is approximately 3.2.

Explanation:

The dissociation of NH4Cl in water can be represented as follows:

NH4Cl(s) → NH4+(aq) + Cl-(aq)

The NH4+ ion can act as a weak acid by donating a proton (H+) to water to form the hydronium ion (H3O+):

NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)

The equilibrium constant expression for this reaction is:

Ka = [NH3][H3O+] / [NH4+]

Since NH3 is a weak base, we can assume that its concentration is negligible compared to that of NH4+. Therefore, we can simplify the expression to:

Ka ≈ [H3O+]^2 / [NH4+]

Substituting the values given in the problem, we get:

1.8 × 10^-5 = [H3O+]^2 / 0.036

Solving for [H3O+], we get:

[H3O+] = √(1.8 × 10^-5 × 0.036) ≈ 6.1 × 10^-4 M

The pH of the solution can be calculated as:

pH = -log[H3O+] ≈ 3.2

User Roland Weisleder
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