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tia_tia [17]
3 years ago
7

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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HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.

<h3>What are buffer solutions and how do they differ?</h3>
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<h3>Describe buffer solution via an example.</h3>
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If 17.4 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentra
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An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h
vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base  =  w

Height of the pyramid = h

Volume of the pyramid = V=\frac{1}{3}lwh

We have:

Rate at which water is filled in cube = \frac{dV}{dt}= 45 cm^3/s

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

V=\frac{1}{3}\times l^2\times h

\frac{l}{h}=\frac{6}{13}

l=\frac{6h}{13}

V=\frac{1}{3}\times (\frac{6h}{13})^2\times h

V=\frac{12}{169}h^3

Differentiating V with respect to dt:

\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}

\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}

Putting, h = 4 cm

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}

=13.20 cm/s

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0
3 years ago
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