A chemist titrates of a aniline solution with HCl solution at . Calculate the pH at equivalence. The of aniline is . Round your answer to decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.

The pH at equivalence is 5.24 .

Explanation:

Let us calculate -

millimoles of aniline = 160 x 0.3403 = 54.448

To reach equivalence , 54.448 millimoles HNO3 must be added,

V = 1086.79 mL HNO3 must be added

total volume = 1086.79 + 160 = 1246.79 mL

Concentration of [salt] = = 0.044 M

Now, at the equivalence or equivalent point ,

By solving the above equation , we get -

pH = 5.24

Hence , the answer is 5.24

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8