What does the pKa of the conjugate acid of a base have to be to remove a hydrogen from water (pKa = 15.7) so that the ratio of h
1 answer:
Answer:
The pKa of the conjugate acid is 17.7
Explanation:
If hydrogen is removed from water, the equilibrium concentration of the conjugate acid according to the information given in the question becomes,
Kₐ = [OH⁻]/[H₂O]
Now, we determine the equivalent pKa
pKa = -log[ka]
pKa = -log[100]
pKa = -2
Removal of hydrogen from water is reversible as shown below;
H₂O ⇄ OH⁻ + H⁺
15.7 -2
This reaction is reversible, and the difference in pKa = pKa[H₂O] - pKa[H⁺];
pKa of the conjugate acid = 15.7 - (-2) = 17.7
The pKa of the conjugate acid is 17.7
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Answer:
650 mmol.
Explanation:
The equation for the fermentation of one mole of glucose is:
C₆H₁₂O₆ + 2 NAD⁺ + 2 ADP + 2 P i + 2 NADH → 2 EtOH + 2 ATP + 2 NADH + 2 NAD⁺
Since NAD⁺/NADH is used and regenerated, we can eliminate it from the equation:
C₆H₁₂O₆ + 2 ADP + 2 P i → 2 EtOH + 2 ATP
With the equation, we calculate the maximum amount of ethanol that could be obtained theoretically:
1000 mmol C₆H₁₂O₆ ------------ 2000 mmol EtOH
325 mmol C₆H₁₂O₆ ------------- x= 650 mmol EtOH
Therefore, the maximum amount of ethanol that could be produced is 650 mmol.
