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.

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.