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NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.084 M in NH4Cl at 25 °C?

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Dejanras
Answer is: pH of solution is 5,17.
Kb(NH₃) = 1,8·10⁻⁵.
c(NH₄Cl) = 0,084 M = 0,084 mol/L.
Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.
Ka · Kb = 10⁻¹⁴.
Ka(NH₄⁺) = 10⁻¹⁴ ÷ 1,8·10⁻⁵.
Ka(NH₄⁺) = 5,55·10⁻¹⁰.
[H₃O⁺] = [NH₃]  = x.
Ka(NH₄⁺) = [H₃O⁺] · [NH₃] ÷ [NH₄⁺].
5,55·10⁻¹⁰ = x² ÷ (0,084 M - x).
Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.
pH = -log[H₃O⁺].
pH = -log(6,8·10⁻⁶ M) = 5,17.
lhmarianateixeira

In this exercise we have to use the ph knowledge of a solution to find its value, so we have:

pH of solution is 5,17.

Organizing the given information as:

  • [tex]Kb(NH_3) = 1,8*10^{-5[/tex]
  • [tex]c(NH_4Cl) = 0,084 M = 0,084 mol/L.[/tex]
  • [tex]NH_4^+ + H_2O \rightarrow NH_3 + H_3O^+[/tex]

Solving the solution ph calculations we find that:

[tex]Ka * Kb = 10^{-14}\\Ka(NH_4^+) = 5,55*10^{-15}\\5,55*10^{-10} = x^2 / (0,084 M - x)\\pH = -log[H_3O^+]\\pH = -log(6,8*10^{-6} M) = 5,17[/tex]

See more about ph at brainly.com/question/15289741

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