The degree to which a weak base dissociates is given by the base-ionization constant, kb. for the generic weak base, b b(aq)+h2o(l)⇌bh+(aq)+oh−(aq) this constant is given by kb=[bh+][oh−][b] strong bases will have a higher kb value. similarly, strong bases will have a higher percent ionization value. percent ionization=[oh−] equilibrium[b] initial×100% strong bases, for which kb is very large, ionize completely (100%). for weak bases, the percent ionization changes with concentration. the more dilute the solution, the greater the percent ionization. ammonia, nh3, is a weak base with a kb value of 1.8×10−5.parta what is the ph of a 0.245 m ammonia solution? partb what is the percent ionization of ammonia at this concentration?

pH of a 0,245 M ammonia solution is 11,3 and percent ionization is 0,86%

Explanation:

For the equilibrium buffer of NH₃:

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻; kb = 1,8x10⁻⁵

kb = [NH₄⁺] [OH⁻] / [NH₃] (1)

When 0,245 M of NH₃ is added, the equilibrium concentrations are:

[NH₃] = 0,245 - x

[NH₄⁺] = x

[OH⁻] = x.

Replacing this values in (1)

x² + 1,8x10⁻⁵x - 4,41x10⁻⁶ = 0

Solving for x:

x = -0,00211 No physical sense. There are not negative concentrations.

x = 0,00211 Real answer

Thus [OH⁻] in equilibrium is 0,00211 M.

As pOH = -log [OH⁻] and 14 = pH + pOH

pH of 0,00211 M is 11,3

It is possible to calculate the percent ionization thus:

Percent ionization = [OH−] equilibrium / [B] initial×100%

Replacing:

0,00211 / 0,245 × 100 = 0,86%

I hope it helps!

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