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Chemical reaction: C₅H₅NH⁺ + H₂O → C₅H₅N + H₃O⁺
c(C₅H₅NH⁺) = 0,21 M = 0,21 mol/dm³.
Kb(pyridine - C₅H₅N) = 1,5·10⁻⁹.
Ka(C₅H₅NH⁺) = 10⁻¹⁴ / 1,5·10⁻⁹.
Ka(C₅H₅NH⁺) = 6,67·10⁻⁶.
c(C₅H₅N) = c(H₃O⁺) = x.
Kb = c(C₅H₅N) · c(H₃O⁺) / c(C₅H₅NH⁺).
6,67·10⁻⁶ = x² / (0,21 M - x).
Solve quadratic equation: x = c(H₃O⁺) = 1,1·10⁻³ mol/dm³.
pH = - logc(H₃O⁺) = -log(1,1·10⁻³ mol/dm³).
pH = 2,9.
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Answer:

pH = 2.9.

Explanation:

What is the ph of a 0.21 m solution of pyridinium chloride (c5h5nh+cl−)? (kb for pyridine is 1.5 × 10−9 .) answer in units of ph?

PH  is the degree of acidity and POH is the measure of  alkalinity

Chemical reaction: C₅H₅NH⁺Cl- + H₂O → C₅H₅N + H₃O⁺

c(C₅H₅NH⁺) = 0,21 M = 0,21 mol/dm³.

Kb(pyridine - C₅H₅N) = 1,5·10⁻⁹.

Ka(C₅H₅NH⁺) = 10⁻¹⁴ / 1,5·10⁻⁹.

Ka(C₅H₅NH⁺) = 6,67·10⁻⁶.

c(C₅H₅N) = c(H₃O⁺) = x.

Kb = [C₅H₅N] · [H₃O⁺] / [C₅H₅NH⁺]

6,67·10⁻⁶ = x² / (0,21 M - x).

factorizing the equation , we have: x = [H₃O⁺] = 1.1x10⁻³ mol/dm³.

pH = - log[H₃O⁺] =

-log(1,1·10⁻³ mol/dm³).

-log(1.1+)-log10⁻³

-log1.1+3

pH = 2.9.

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