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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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Cyclohexane (C6H12) is a hydrocarbon (a substance containing only carbon and hydrogen) liquid. Which of the following will most
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Now as it is a non polar species thus it will dissolve only a non polar solute following the principle of 'Like dissolves Like'. Thus the most non polar species given in the options is pentane. Cyclohexane will definitely dissolve this pentane as both are non polar in nature.

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The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react
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<u>Answer:</u> The value of K_{eq} is 4.84\times 10^{-5}

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}

So, concentration of ammonia = \frac{0.0280}{1.00}=0.0280M

Concentration of oxygen gas = \frac{0.0120}{1.00}=0.0120M

The given chemical equation follows:

                  4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)

<u>Initial:</u>        0.0280        0.0120

<u>At eqllm:</u>    0.0280-4x   0.0120-3x   2x       6x

We are given:

Equilibrium concentration of nitrogen gas = 3.00\times 10^{-3}M=0.003

Evaluating the value of 'x', we get:

\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M

Now, equilibrium concentration of ammonia = 0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M

Equilibrium concentration of oxygen gas = 0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M

Equilibrium concentration of water = 6x=(6\times 0.0015)]=0.009M

The expression of K_{eq} for the above reaction follows:

K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}

Putting values in above expression, we get:

K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}

Hence, the value of K_{eq} is 4.84\times 10^{-5}

4 0
3 years ago
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