Question

1. Consider the titration of 0.250 M of 50.0 mL nitric acid (HNO) by 0.350 M NH,. Kb for NH, is 1.8 x 10°. Calculate the pH of the resulting solution after the following volumes of NH, have been added. a. 0.0 mL b. 50.0 mL c. Half-way equivalence point d. At equivalence point e. 150.0 mL​

Answer 1

Answer:I got pH = 4.98

Explanation:. This agrees with the fact that strong acid + weak base = acidic pH.

If you notice, the volume and concentration of both

NH

3

and

HNO

3

are identical, and

HNO

3

only has one

H

+

. Therefore, you should convince yourself that these neutralize each other exactly because the number of mols of each are equal.

This leaves the same number of mols of

NH

+

4

as

NH

3

, in TWICE the volume (i.e. don't forget to account for dilution!):