As the sand mound grows, the point of maximum sand deposition on the leeward face moves closer to the summit, causing a steepening of the leeward face relative to the windward face. The steepening and growing dune now forces wind out over the top of the dune rather than down the leeward face.

7 0

3 years ago

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

Colt1911 [192]

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!

6 0

2 years ago