Answer:
24.8 molecules are ionized from 1000 acetic acid molecules.
Explanation:
Acetic acid, CH₃COOH dissociates in water, thus:
CH₃COOH ⇄ CH₃COO⁻ + H⁺
Ka = 6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]
<em>That means amount of CH₃COO⁻ (the dissociated form) that are produced is followed by the equilibrium of the weak acid.</em>
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The initial molar concentration of acetic acid (Molar mass: 60g/mol) is:
6g ₓ (1mol / 60g) = 0.1 moles acetic acid, in 1000cm³ = 1L.
0.1 moles / L = <em>0.1M</em>
The 0.1M of acetic acid will dissociate producing X of CH₃COO⁻ and H⁺, thus:
[CH₃COOH] = 0.1M - X
[CH₃COO⁻] = X
[H⁺] = X
Replacing in Ka formula:
6.3x10⁻⁵ = [CH₃COO⁻] [H⁺] / [CH₃COOH]
6.3x10⁻⁵ = [X] [X] / [0.1 - X]
6.3x10⁻⁶ - 6.3x10⁻⁵X = X²
6.3x10⁻⁶ - 6.3x10⁻⁵X - X² = 0
Solving for X
X = - 0.0025 → False solution, there is no negative concentrations.
X = 0.00248M
That means, a 0.1M of acetic acid produce:
[CH₃COO⁻] = X = 0.00248M solution of the ionized form.
In a basis of 1000 molecules:
1000 molecules × (0.00248M / 0.1M) = 24.8
<h3>24.8 molecules are ionized from 1000 acetic acid molecules.</h3>
m(H2SO4) which is =<span>470g