A chemist titrates 100.0 mL of a 0.3065 M aniline (C6H5NH2) solution with 0.5774 solution at 25 . Calculate the pH at equivalence. The of aniline is .

4.91

Explanation:

From the given information:

The millimoles of aniline = 100* 0.3065 = 30.65

30.65 millimoles HNO3 must be added to reach the equivalence point.

30.65 = V x 0.5774

V = 30.65/0.5774

V = 53.08 mL HNO3 must be added

Thus, total volume = 53.08 + 100 = 153.08 mL

[salt] = 30.65 /  153.08 = 0.20022 M

At equivalence point :

pOH = 1/2 [pKw + pKb + logC]

pOH = 1/2 [14 + 4.87 + log 0.20022 ]

pOH =9.086

pH = 14 - 9.086

pH = 4.914

pH ≅ 4.91 to 2 decimal places.

there are two significant figures.