Question 1

How do pKa values determine which side of an acid-base equilibrium is favored?

Accepted Answer

The equilibrium position in an acid-base reaction favors formation of the weaker acid and weaker base because they are more stable species with lower potential energies. By comparing pKa values, the stronger acid (lower pKa) is identified. The reaction shifts toward products containing the weaker acid (higher pKa) and its conjugate base. For example, ethanol with pKa 15.9 is stronger than ammonia with pKa 38, so equilibrium favors ammonia and ethoxide ion formation.

Question 2

Why does a strong acid always produce a weak conjugate base?

Accepted Answer

Strong acids readily donate protons, leaving behind conjugate bases with minimal tendency to accept protons back. Conversely, weak acids hold protons more tightly, producing conjugate bases that strongly attract protons. This inverse relationship means the stronger the acid, the weaker its conjugate base, and vice versa. In the ammonia-ethoxide reaction, ethanol is the stronger acid, making ethoxide ion the weaker base compared to amide ion.

Question 3

What does the equilibrium constant value tell you about reaction direction?

Accepted Answer

The equilibrium constant is calculated by subtracting the pKa of the stronger acid from the pKa of the weaker acid, then taking the antilog. When Keq is greater than 1, equilibrium lies far toward products. When Keq is less than 1, equilibrium lies far toward reactants. A large Keq indicates the reaction strongly favors product formation, reflecting the stability difference between reactants and products.

Question 4

When should reversible arrows be replaced with irreversible arrows in acid-base reactions?

Accepted Answer

If the difference in pKa values between the two acids is large, the reversible reaction is negligible, and reversible arrows are replaced with a single irreversible arrow. This indicates the reaction goes essentially to completion in one direction. The larger the pKa difference, the more completely the reaction favors one side, making the reverse reaction insignificant under standard conditions.

Question 5

How can you predict equilibrium position using the pKa difference between acids?

Accepted Answer

Subtract the pKa of the weaker acid from the pKa of the stronger acid and take the antilog to find the equilibrium constant. For the ammonia-ethoxide reaction, pKa(ammonia) minus pKa(ethanol) equals 38 minus 15.9, giving log Keq of 22.1. The antilog yields a very large Keq, confirming equilibrium strongly favors the side with the weaker, more stable acid and relative stability and degree of solvation factors.

Question 6

What role does potential energy play in determining equilibrium position?

Accepted Answer

Acid-base equilibria favor formation of species with lower potential energies because they are thermodynamically more stable. Weaker acids and weaker bases possess lower potential energies than their stronger counterparts. This fundamental principle explains why equilibrium shifts toward products containing the weaker acid and its conjugate base, regardless of reaction conditions or mechanism.

Question 7

In the ammonia-ethoxide reaction, why does equilibrium favor ammonia and ethoxide ion formation?

Accepted Answer

Ethanol (pKa 15.9) is a stronger acid than ammonia (pKa 38), making ammonia the weaker acid. Since strong acids form weak conjugate bases, ethoxide ion is a weaker base than amide ion. Equilibrium favors formation of the weaker acid (ammonia) and weaker base (ethoxide ion) because these species have lower potential energies and greater stability.

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