Butane (c4h10) + oxygen -> carbon dioxide + water
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Answer:
The solution 0.010 M has a higher percent ionization of the acid.
Explanation:
The percent ionization can be found using the following equation:
Since we know the acid concentration in the two cases, we need to find [H₃O⁺].
By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:
1. Case 1 (0.1 M):
CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq) (1)
0.1 - x x x
(2)
Where:
Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.
By solving the above equation for x we have:
x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]
Hence, the percent ionization is:
2. Case 2 (0.01 M):
The dissociation constant from reaction (1) is:
With [CH₃COOH] = 0.01 M
By solving the above equation for x:
x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]
Then, the percent ionization for this case is:
As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.
Therefore, the solution 0.010 M has a higher percent ionization of the acid.
I hope it helps you!