Question

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?

Answer 1

Answer:

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)

$1,8x10^{-5} = \frac{x^2}{0,245 - x}$

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!