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!

3

explanation:

plants need sunlight, water, and carbon dioxide to make food. plants use the raw materials of water, nutrients they absorb from soil, and carbon dioxide to make sugars the plant can use. energy from the sun and chlorophyll are also needed.