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Brrunno [24]
2 years ago
6

Explain why the dissociation of a weak acid can be ignored when calculating the pH of a solution that contains both a weak acid

and a strong acid. The ion from the strong acid is a common ion in the dissociation reaction of the weak acid. Consequently, it shifts the equilibrium back toward the which the dissociation of the to the point where the amount of ion that it produces is negligible
Chemistry
1 answer:
Sliva [168]2 years ago
4 0

Answer:

The answer is in the explanation.

Explanation:

The dissociation of a weak acid consist in the following equilibrium:

HX ⇄ H⁺ + X⁻

Where Ka is defined as:

Ka =  [H⁺] [X⁻] / [HX]

A strong acid (HY) dissociates completely in water, thus:

HY → H⁺ + Y⁻

As the strong acid produces H⁺, in the equilibrium, the reaction shifts to the left -The undissociated form-, reducing the production of H⁺, allowing ignore the dissociation of the weak acid when calculating the pH.

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a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill
baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

PV=nRT

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R is Gas constant having value = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Also,  

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So, the ideal gas equation can be written as:

PM=dRT

Given that:-

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The conversion of T( °C) to T(K) is shown below:

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So,  

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Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

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