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ELEN [110]
3 years ago
9

In a titration of monoprotic acids and bases, there is a large change in pHA. at the point where pH = pKa of the acid.B. when th

e volume of acid is exactly equal to the volume of base.C. when the concentration of acid is exactly equal to the concentration of base.D. when the number of moles of acid is exactly equal to the number of moles of base.E. at the point where pH = pKb of the
Chemistry
1 answer:
Jobisdone [24]3 years ago
3 0

Answer:

D. when the number of moles of acid is exactly equal to the number of moles of base.

Explanation:

<em>Regarding options A. and E</em>., pKa and pKb would only be taken into consideration if the titrations were of <em>weak</em> acids and bases. However it is possible to have a titration of monoprotic acids and bases with strong acids and bases.

Another way of looking at the answer is identifying <em>which one best describes the equivalence point</em>.

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The standard free-energy changes for the reactions below are given. Phosphocreatine → creatine + Pi ∆ G'° = –43.0 kJ/mol ATP → A
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Answer:

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Explanation:

The free-energy predicts if a reaction is spontaneous or not. If it is, ΔG < 0. When a reaction happens by steps, the free-energy of the global reaction can be calculated by the sum of the free-energy of the steps (Hess law). If it's needed to operations at the reaction the same operation must be done in the value of ΔG (if the reaction is inverted, the signal of ΔG must be inverted).

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

ATP → ADP + Pi                             ∆G'° = –30.5 kJ/mol (x-1)

--------------------------------------------------------------------------------------

Phosphocreatine → creatine + Pi ∆G'° = –43.0 kJ/mol

Pi + ADP → ATP                             ∆G'° = 30.5 kJ/mol

The bold compounds are in opposite sides, so they'll be canceled in the sum of the reactions:

Phosphocreatine + ADP → creatine + ATP

∆G'° = -43.0 + 30.5

∆G'° = -12.5 kJ/mol

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