Carbonic acid (H₂CO₃) is a polyprotic acid. When carbonic acid dissolves in water, which is higher, the concentration of HCO₃- i
1 answer:
The concentration of CO₃²⁻ ions will be higher
To explain, I want you to imagine H₂CO₃ in water.
we know that it will lose 2 of it's protons, and form 2 ions
The ion which is more stable will have a higher concentration because that ion will refuse to react with anything else, so once something turns into that specific ion, it's not going back... unless there's a more stabler ion possible
In this case, the 2 ions formed are: HCO₃⁻ and CO₃⁽²⁻⁾, drawing the structures of both the ions tells us that both of them have resonance, but the CO₃⁽²⁻⁾ ion has more resonance structures and hence is more stable
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Answer:
0.013%
Yes, it does. The answer agrees with the statement.
Explanation:
Both conformers are in equilibrium, and it can be represented by the equilibrium equation K:
K = [twist-boat]/[chair]
The free energy between them can be calculated by:
ΔG° = -RTlnK
Where R is the gas constant (8.314 J/mol.K), and T is the temperature (25°C + 273 = 298 K).
ΔG° = 5.3 kcal/mol * 4.182 kJ/kcal = 22.165 kJ/mol = 22165 J/mol
22165 = -8.314*298*lnK
-2477.572lnK = 22165
lnK = -8.946
K =
K = 1.30x10⁻⁴
[twist-boat]/[chair] = 1.30x10⁻⁴
[twist-boat] = 1.30x10⁻⁴[chair]
The percentage of the twist-boat conformer is:
[twist-boat]/([twist-boat] + [chair]) * 100%
1.30x10⁻⁴[chair]/(1.30x10⁻⁴[chair] + [chair]) *100%
0.013%
The statement about the conformers is that the chair conformer is more stable, and because of that is more present. So, the answer agrees with it.