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2 years ago

Consider molecular, complete ionic, and net ionic equations.

Chemistry

1 answer:

8090 [49]2 years ago

3 0

Explanation:

(a) The molecular equations shows the equation in which all the species of the reactants and the products are in molecules and the net charge is zero.

The complete ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and are represented as ions.

The net ionic equations shows the equation in which all the species of the reactants and the products are in dissociated form and do not show the spectator ions which are same in the reactants and the products.

(b) If there is no spectator ions in the reaction, then the complete and the net ionic equations would be identical.

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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%

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3 years ago