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

answer: i have one answer

acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (h+) ions

explanation:

hope this and sorry i could only come up with one answer!

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.