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alonzomacias1989
25.02.2020 •
Chemistry
The pH of a 1.4M solution of barbituric acid HC4H3N2O3 is measured to be 1.93. Calculate the acid dissociation constant Ka of barbituric acid. Round your answer to 2 significant digits.
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Ответ:
9.8 × 10⁻⁵
Explanation:
Let's consider the acid dissociation of barbituric acid.
HC₄H₃N₂O₃(aq) ⇄ C₄H₃N₂O₃⁻(aq) + H⁺(aq)
The pH is 1.93. We can find the concentration of H⁺.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -1.93 = 0.0117 M
The concentration of the acid (Ca) is 1.4 M. We can fidn the acid dissociation constant (Ka) using the following expression.
[H⁺] = √(Ca × Ka)
[H⁺]² = Ca × Ka
Ka = [H⁺]² / Ca
Ka = (0.0117)² / 1.4
Ka = 9.8 × 10⁻⁵
Ответ:
————
x+2