A chemist titrates 190.0 mL of a 0.8125 M ammonia (NH) solution with 0.3733 M HCl solution at 25 °C. Calculate the pH at equivalence. The pK, of ammonia is 4.75.
Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HCl solution added.
pH ?

Approximately .

Explanation:

Initial volume of the solution: .

Initial quantity of :

.

Ammonia reacts with hydrochloric acid at a one-to-one ratio:

.

Hence, approximately of molecules would be required to exactly react with the in the original solution and hence reach the equivalence point of this titration.

Calculate the volume of that solution required for reaching the equivalence point of this titration:

.

Hence, by the assumption stated in the question, the volume of the solution at the equivalence point would be approximately .

If no hydrolysis took place, of would be produced. Because is a soluble salt, the solution would contain of ions. The concentration of would be approximately:

.

However, because is a weak base, its conjugate would be a weak base.

.

Hence, the following reversible reaction would be take place in the solution at the equivalence point:

.

Let be the increase in the concentration of in this solution because of this reversible reaction. (Notice that .) Construct the following table:

.

Thus, at equilibrium:

.

Solve for . (Notice that the value of is likely to be much smaller than . Hence, the denominator on the left-hand side .)

.

Hence, the concentration of at the equivalence point of this titration would be approximately .

Hence, the at the equivalence point of this titration would be:

.

d

explanation: