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3 years ago

If a compound contains an anion which is the conjugate base of a weak acid, the addition of hydronium ion solubility does what ?

Chemistry

1 answer:

wel3 years ago

3 0

Answer:

The addition of hydronium ion increases the solubility of the salt

Explanation:

When a weak acid, MA dissociates in an aqueous solution, the following products in equilibrium are obtained;

MA ----> M+ + A-

The anion which is a conjugate base will be removed from the solution with the addition of a hydronium ion from an acid: 2H3O+ + 2A- ----> 2HA + 2H20

This distorts the reaction equilibrium. According to Le Chatelier’s principle, more MA will dissolve until in order to restore the previous equilibrium of the reaction.

Therefore, an acidic pH increases the solubility of almost all sparingly soluble salts whose anion is the conjugate base of a weak acid.

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A sample of nitrogen gas has a temperature of 22.7oC when the volume of the container is 12.2L and it is under 150.4kPa of press

Marrrta [24]

Answer:

The final temperature of the gas would need to be approximately 158.4 K

Explanation:

The details of the sample of nitrogen gas are;

The initial temperature of the nitrogen gas, T₁ = 22.7°C = 295.85 K

The initial volume occupied by the gas, V₁ = 12.2 L

The initial pressure of the gas, P₁ = 150.4 kPa

The final volume of the gas, V₂ = 9.7 L

The final pressure of the gas, P₂ = 101.3 kPa

Let 'T₂', represent the final temperature of the gas, by the ideal gas equation, we have;

$\dfrac{P_1 \times V_1}{T_1} = \dfrac{P_2 \times V_2}{T_2}$

$\therefore \ T_2 = \dfrac{P_2 \times V_2 \times T_1 }{P_1 \times V_1}$

Plugging in the values gives;

$\therefore \ T_2 = \dfrac{101.3 \, kPa \times 9.7 \ L \times 295.85 K}{150.4 \ kPa \times 12.2 \ L} \approx 158.4327959 \ K$

The final temperature of the gas, T₂ ≈ 158.4 K

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KengaRu [80]

Answer:

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GuDViN [60]

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8 0

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A generic salt, ab2, has a molar mass of 169 g/mol and a solubility of 1.10 g/l at 25 °c. what is the ksp of this salt at 25 °c?

GaryK [48]

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Ksp=[A²⁺][B⁻]²

a=1.10 g/L
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c=a/M

[A²⁺]=c=a/M
[B⁻]=2c=2a/M

Ksp=(a/M)(2a/M)²=4(a/M)³

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