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A) If Kb for NX3 is 4.5×10^−6, what is the pOH of a 0.175 M aqueous solution of NX3 ?

B) If Kb for NX3 is 4.5×10^−6, what is the percent ionization of a 0.325 M aqueous solution of NX3 ?

Please show your work so I can understand how to solve similar problems on my own! Thank you!

Respuesta :

3.08 is the pOH of a 0.175 M aqueous solution of [tex]NX_3[/tex].

0.215% is the per cent ionization of a 0.325 M aqueous solution of [tex]NX_3[/tex]

What is pH?

pH is a measure of how acidic/basic water is.

A)

[tex]NX_3 + H_2O[/tex] →[tex]NHX_3^+ + OH^-[/tex]

Kb = 4.5 x[tex]10^-6[/tex]

Kb = {concentration of (NH₄⁺) x concentration of (OH⁻)} ÷ concentration of (NH₃).

concentration of (NH₄⁺) = concentration of (OH⁻) = x.

= Kb x concentration of (NH₃)

= 4.5 × 10⁻⁶ × 0.175 = 7.0 × 10⁻⁷.

x = concentration of (OH⁻) = √(7.0 × 10⁻⁷)

= 8.367 × 10⁻⁴

pOH = -log(c(OH⁻))

=- log ( 8.367 × 10⁻⁴)

= 3.08

B)

Chemical reaction: NX₃ + H₂O ⇄ NX₃H⁺ + OH⁻.

Concentration of (NX₃) = 0.325 M.

Kb = 4.5 x 10⁻⁶.

[NX₃H⁺] = [OH⁻] = x.

[NX₃] = 0.325 M - x.

Kb = [NX₃H⁺] x [OH⁻] ÷  [NX₃].

4.5 x 10⁻⁶ = x² ÷ (0.325 M - x).

x = 0.0007 M.

Per cent of ionization:

α = 0. 0007 M ÷ 0. 325 M x 100%

= 0.215%.

Hence,

3.08 is the pOH of a 0.175 M aqueous solution of [tex]NX_3[/tex].

0.215% is the per cent ionization of a 0.325 M aqueous solution of [tex]NX_3[/tex]

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https://brainly.com/question/12353627

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