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IRINA_888 [86]
4 years ago
6

Which of the following statements concerning equilibrium constants is true? The equilibrium constant of the reverse reaction is

equal in magnitude but opposite in sign of the forward reaction. When related reactions are added together, their equilibrium constants are added together. When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is multiplied by two. When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is not affected. When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is squared.
Chemistry
1 answer:
Brut [27]4 years ago
3 0

Answer:

When the coefficients in a balanced chemical reaction are multiplied by two, the equilibrium constant is not affected.

Explanation:

The equilibrium constant of a reaction is known to remain steady.

Even if all the coefficients of all the species in the reaction are multiplied by two, the value of the equilibrium constant will reamin the same because the equilibrium position will not change as a result of that.

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Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

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The painkiller, Advil® contains the active ingredient ibuprofen (IB), which has a pKb of
denis23 [38]

This problem is providing the basic dissociation constant of ibuprofen (IB) as 5.20, its pH as 8.20 and is requiring the equilibrium concentration of the aforementioned drug by giving the chemical equation at equilibrium it takes place. The obtained result turned out to be D) 4.0 × 10−7 M, according to the following work:

First of all, we set up an equilibrium expression for the given chemical equation at equilibrium, in which water is omitted for it is liquid and just aqueous species are allowed to be included:

Kb=\frac{[IBH^+][OH^-]}{[IB]}

Next, we calculate the concentration of hydroxide ions and the Kb due to the fact that both the pH and pKb were given:

pOH=14-8.20=5.80

[OH^-]=10^{-5.8}=1.585x10^{-6}M

Kb=10^{-5.20}=6.31x10^{-6}

Then, since the concentration of these ions equal that of the conjugated acid of the ibuprofen (IBH⁺), we can plug in these and the Kb to obtain:

6.31x10^{-6}=\frac{(1.585x10^{-6})(1.585x10^{-6})}{[IB]}

Finally, we solve for the equilibrium concentration of ibuprofen:

[IB]=\frac{(1.585x10^{-6})(1.585x10^{-6})}{6.31x10^{-6}}=4.0x10^{-7}

Learn more:

(Weak base equilibrium calculation) brainly.com/question/9426156

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