You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.18. You determine that the concent
1 answer:
Answer:
Explanation:
Hello,
In this case, given the information, we can compute the concentration of hydronium given the pH:
Next, given the concentration of the acid and due to the fact it is monoprotic, its dissociation should be:
We can write the law of mass action for equilibrium:
Thus, due to the stoichiometry, the concentration of hydronium and A⁻ are the same at equilibrium and the concentration of acid is:
As the concentration of hydronium also equals the reaction extent (). Thereby, the acid dissociation constant turns out:
And the pKa:
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Answer:
Explanation:
2NO₂ ⇌ N₂O₄
E/mol·L⁻¹: 0.058 0.012
K_{\text{eq}} = \dfrac{\text{[N$_{2}$O$_{4}$]}}{\text{[NO$_{2}$]$^{2}$}} = \dfrac{0.012}{0.058^{2}} = \mathbf{3.6} \\\\
\text{The $K_{\text{eq}}$ value would be $\boxed{\mathbf{3.6}}$}