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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

Respuesta :

Answer:

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

[tex]Molarity\ of\ HI = \frac{moles}{volume \ in\ L}[/tex]

[tex]= \frac{4.63\times10^{-2}}{0.250}[/tex]

= 0.1852  M

Now

As we know that

[tex]NH_3 + HI = NH_4I[/tex]

So,

[tex]NH_4I = HI = 0.1852 M[/tex]

Now the molarity of [tex]NH_3[/tex] left is

= 0.397 - 0.1852

= 0.2118

[tex]pOH = pKb + log (\frac{NH_4I}{NH_3})[/tex]

[tex]= 4.75 + log(\frac{0.1852}{0.2118})[/tex]

= 4.692

Now as we know that

pH = 14 - pOH

= 14 - pOH

= 14 - 4.692

= 9.308

We simply applied the above equations

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