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

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