Question

a)Explain why it is important that the equilibrium constant for the reaction that occurs during an acid-base titration be very large.

b)

a)Explain why it is important that the equilibrium constant for the reaction that occurs during an acid-base titration be very large. b) Microscopic representations for four different solutions containing bases are shown below. What are the relative strengths of the bases shown in l - IV? Explain your reasoning. Note: H2O molecules are not shown for clarity.

Answer 1

a)

Reactants --------->       Products
       k = P/R =   Products/reactants
so a higher equillibirum constant means more products. More products means, the reaction is favored in forward direction.

If the value for K_c is very large, then the equilibrium favors the reaction to the right and there are more products than reactants. The reaction may be said to be "complete" or "quantitative."

If the value for the equilibrium constant is small, then the equilibrium favors the reaction to the left and there are more reactants than products. If the value of K_c approaches zero the reaction may be considered not to occur.

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b)

The reaction that is involving here in this is

B + H2O <------------>  BH+ + OH-

Base molecule gets dissociated as BH+ and OH- ions.

Let us represent each diagram in terms of chemical equation. All the solutions are formed from dissociation of 4B molecules. They are represented as

I : 4OH- + 4BH+

As, this undergo complete dissociation, this is the strongest base.

II: 3B + OH- + BH+

This is the solution which undergoes least dissociation leaving 3B out. So, this is the weakest base.

III: 2B + 2OH- + 2BH+

This is the second least weakest base.

IV: B + 3OH- + 3BH+

This undergoes almost complete dissociation, but one B is left out. So, this is the second strongest base.

Thus, the order of solutions of relative strengths of bases is

I > IV > III > II