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prue10
@prue10
January 2021
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Bonjour
C'est. Sur les acides faibles et bases faibles
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scoladan
Verified answer
Bonjour,
1) acide base conjuguée
AH A-
----------------------|-------------------------->
pKa = 3,8
2) pH = 6,5
donc pH > pKa
⇒ La forme basique (ions lactate) prédomine.
3) [H₃O⁺] = 10^(-pH) = 10^(-6,5) ≈ 3,16.10⁻⁷ mol.L⁻¹
4)
AH + H₂O = A⁻ + H₃O⁺
Ka = [A⁻] x [H₃O⁺]/[AH]
5) pKa = -log(Ka)
⇒ Ka = 10^(-pKa) = 10^(-3,8) ≈ 1,58.10⁻⁴
et donc : [A⁻]/[AH] = Ka/[H₃O⁺] = 1,58.10⁻⁴/3,16.10⁻⁷ ≈ 501
Donc [A⁻] >> [AH]
ce qui confirme la réponse à la question 2
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Lista de comentários
Verified answer
Bonjour,1) acide base conjuguée
AH A-
----------------------|-------------------------->
pKa = 3,8
2) pH = 6,5
donc pH > pKa
⇒ La forme basique (ions lactate) prédomine.
3) [H₃O⁺] = 10^(-pH) = 10^(-6,5) ≈ 3,16.10⁻⁷ mol.L⁻¹
4)
AH + H₂O = A⁻ + H₃O⁺
Ka = [A⁻] x [H₃O⁺]/[AH]
5) pKa = -log(Ka)
⇒ Ka = 10^(-pKa) = 10^(-3,8) ≈ 1,58.10⁻⁴
et donc : [A⁻]/[AH] = Ka/[H₃O⁺] = 1,58.10⁻⁴/3,16.10⁻⁷ ≈ 501
Donc [A⁻] >> [AH]
ce qui confirme la réponse à la question 2