A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equivalence. The pKb of aniline is 4.87 . 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 HNO3 solution added.

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of

Formula used :

where,

are the initial molarity and volume of .

are the final molarity and volume of .

We are given:

Putting values in above equation, we get:

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of +  Volume of

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

Now we have to calculate the pH of the solution.

At equivalence point,

Thus, the pH of the solution is, 5.24

gold is my favorite solid.