An aqueous solution contains 0.397 M ammonia. Calculate the pH of the solution after the addition of 4.63 x 10-2 moles of perchloric acid (HClO4) to 250 mL of this solution. (Assume the volume does not change upon adding perchloric acid). Ka = 5.7 x 10-10, Kb = 1.80 x 10-5

9.308

Explanation:

The computation of the pH of the given solution is shown below:

But before we need to determine the HI molarity which is

= 0.1852  M

Now

As we know that

So,

Now the molarity of left is

= 0.397 - 0.1852

= 0.2118

= 4.692

Now as we know that

pH = 14 - pOH

= 14 - pOH

= 14 - 4.692

= 9.308

We simply applied the above equations

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong