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BabaBlast [244]
3 years ago
5

A student prepares a 0.47mM aqueous solution of acetic acid CH3CO2H. Calculate the fraction of acetic acid that is in the dissoc

iated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource. Round your answer to 2 significant digits. %
Chemistry
1 answer:
Aliun [14]3 years ago
6 0

The fraction of acetic acid that is dissociated is 0.18

Why?

The chemical equation for the dissociation of acetic acid (HAc) is the following:

HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)

To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:

f=\frac{-Ka+\sqrt{Ka^{2} +4KaC} }{2C} = 0.18

Have a nice day!

#LearnwithBrainly

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The burning of a sample of propane
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Answer:

70.0°C

Explanation:

We are given;

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We are required to determine the final temperature of water;

Taking the initial temperature is x°C

We know that the specific heat of water is 4.18 J/g°C

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Change in temp =(x-20)° C

Therefore;

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The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori
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Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

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