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Anna71 [15]
3 years ago
15

Acetic acid (CH3COOH) is a weak acid that partially dissociates in water. Given the reaction CH3COOH(aq) ↔ CH3COO−(aq) + H+(aq)

and Kc = 1.8 x 10−5, if the concentration of the acetic acid is 0.016 M and the concentration of the acetate ion is 0.92 M, what is the [H+]?
Chemistry
1 answer:
likoan [24]3 years ago
8 0

Answer:

The correct answer is: 1.035 x 10⁻³ M

Explanation:

The dissociation equilibrium for acetic acid (CH₃COOH) is the following:

CH₃COOH(aq) ↔ CH₃COO⁻(aq) + H⁺(aq)  Kc = 1.8 x 10⁻⁵

The expression for the equilibrium constant (Kc) is the ratio of concentrations of products over reactants. The products are acetate ion (CH₃COO⁻) and hydrogen ion (H⁺) while the reactant is acetic acid (CH₃COOH):

Kc=\frac{[CH_{3} COO^{-} ][H^{+} ]}{[CH_{3} COOH]}= 1.8 x 10^{-5}

Given: [CH₃COOH]= 0.016 M and [CH₃COO⁻]= 0.92 M, we replace the concentrations in the equilibrium expression and we calculate [H⁺]:

\frac{(0.016 M)[H^{+} ]}{(0.92M)}= 1.8 x 10^{-5}

⇒[H⁺]= (1.8 x 10⁻⁵)(0.92 M)/(0.016 M)= 1.035 x 10⁻³ M

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Mazyrski [523]

Answer:

Lead shows the greatest temperature change upon absorbing 100.0 J of heat.

Explanation:

Q=mc\Delte T

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ΔT = change in temperature

1) 10.0 g of copper

Q = 100.0 J (positive means that heat is gained)

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\Delta T=\frac{Q}{mc}

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m = 10.0 g

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\Delta T=\frac{Q}{mc}

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Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the water = c =  4.18J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 4.18 J/g^oC}=2.39 ^oC

5) 10.0 g of lead

Q = 100.0 J (positive means that heat is gained)

m = 10.0 g

Specific heat of the lead= c =  0.128 J/g°C

\Delta T=\frac{Q}{mc}

=\frac{100.0 J}{10 g\times 0.128 J/g^oC}=78.125^oC

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