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The pOH of an aqueous solution at 25°C was found to be 8.80 . The pH of this sol

ID: 998856 • Letter: T

Question

The pOH of an aqueous solution at 25°C was found to be8.80.
The pH of this solution is .
The hydronium ion concentration is  M.
The hydroxide ion concentration is  M.

The pH of an aqueous solution at 25°C was found to be 8.50.
The pOH of this solution is .
The hydronium ion concentration is  M.
The hydroxide ion concentration is  M.

What is the pOH of an aqueous solution of 0.323 M perchloric acid?
pOH = M

What is the hydronium ion concentration in an aqueoushydroiodic acid solution with a pH of 1.680?
[H3O+] = M

What is the hydronium ion concentration in an aqueousperchloric acid solution that has a pOH of 9.700?
[H3O+] = M

Explanation / Answer

Answer – 1) We are given, pOH = 8.80

We know,

pH +pOH = 14

so, pH = 14 – pOH

            = 14 – 8.80

            = 5.20

Now hydronium ion

We know,

pH = -log [H3O+]

[H3O+] = 10-pH

            = 10-5.2

             = 6.31*10-6 M

Hydroxide ion –

We know,

pOH = -log [OH-]

[OH-] = 10-pOH

            = 10 -8.80

             = 1.58*10-9 M

2) We are given, pH = 8.50

We know,

pH +pOH = 14

so, pOH = 14 – pH

            = 14 – 8.50

            = 5.50

Now hydronium ion

We know,

pH = -log [H3O+]

[H3O+] = 10-pH

            = 10-8.50

             = 3.16*10-9 M

Hydroxide ion –

We know,

pOH = -log [OH-]

[OH-] = 10-pOH

            = 10 -5.50

             = 3.16*10-6 M

3) We are given the [HClO4] = 0.324 M

We know perchloric acid strong acid, so [HClO4] = [H3O+]

So, [H3O+] = 0.324 M

We know,

pH = -log [H3O+]

     = -log 0.324 M

     = 0.489

We know,

pH +pOH = 14

so, pOH = 14 – pH

            = 14 – 0.489

            = 13.5

4) We are given, pH = 1.680

We know,

pH = -log [H3O+]

[H3O+] = 10-pH

            = 10-1.680

             = 0.0209 M

5) We are given, pOH = 9.700

We know,

pH +pOH = 14

so, pH = 14 – pOH

            = 14 – 9.70

            = 4.3

Now hydronium ion

We know,

pH = -log [H3O+]

[H3O+] = 10-pH

            = 10-4.3

             = 5.01*10-5 M

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