Question

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

Answer 1

9.308

Step-by-step explanation:

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

But before we need to determine the HI molarity which is

`Molarity\ of\ HI = (moles)/(volume \ in\ L)`

`= (4.63*10^(-2))/(0.250)`

= 0.1852 M

Now

As we know that

`NH_3 + HI = NH_4I`

So,

`NH_4I = HI = 0.1852 M`

Now the molarity of
`NH_3` left is

= 0.397 - 0.1852

= 0.2118

`pOH = pKb + log ((NH_4I)/(NH_3))`

`= 4.75 + log((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