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2 years ago

If kb for nx3 is 4.0×10−6, what is the poh of a 0.175 m aqueous solution of nx3?

Chemistry

1 answer:

Elena-2011 [213]2 years ago

7 0

<h3>Answer;</h3>

pOH = 3.08

<h3>Explanation;</h3>

NX3 + H2O <----> NHX3+ + OH-

Kb = 4.0 x 10^-6

Kb = c(NH₄⁺) · c(OH⁻) ÷ c(NH₃).

c(NH₄⁺) = c(OH⁻) = x.

x² = Kb · c(NH₃)

x² = 4.0 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = c(OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

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