Determine the pOH of each solution.

(a) [H3O+] = 8.3 × 10−10 M

(b) [H3O+] = 1.6 × 10−7 M

(c) [H3O+] = 7.3 × 10−2 M

(d) [OH−] = 4.32 × 10−4 M

Solution: Here, we are going to determine the pOH of the given solutions.

Step1:

In pure water, one H2O molecule donates proton and acts as an acid and another water molecule accepts a proton and acts as a base at the same time. The following equilibrium exists:

H2O(l) + H2O(l) ------> H3O+(aq) + OH-(aq)

The equilibrium constant for the reaction is given by:

Kw = [H3O+][OH−] -----(1)

Where, Kw is called the ionic product of water and its value is 1 x 10-14 M2.

moreover, pH + pOH =14

Where, pOH = - log[OH- ] and pH = - log[H+ ]

Step2:

- Given, [H3O+] = 8.3 x 10-10 M

From equation (1), [OH−] = Kw / [H3O+]

[OH−] = (1 x 10-14 M2) / (8.3 x 10-10 M)

[OH−] = 1.2 x 10-5 M

Therefore pOH = - log[OH- ]

= - log (1.2 x 10-5)

= - log 1.2 - log 10-5

= -0.079 + 5

pOH = 4.921

b) Given, [H3O+] = 1.6 x 10-7 M

From equation (1), [OH−] = Kw / [H3O+]

[OH−] = (1 x 10-14 M2) / (1.6 x 10-7 M)

[OH−] = 6.25 x 10-8 M

Therefore pOH = - log[OH- ]

= - log (6.25 x 10-8)

= - log 6.25 - log 10-8

= -0.796 + 8

pOH = 7.204

c) Given, [H3O+] = 7.3 x 10-2 M

From equation (1), [OH−] = Kw / [H3O+]

[OH−] = (1 x 10-14 M2) / (7.3 x 10-2 M)

[OH−] = 1.37 x 10-13 M

Therefore pOH = - log[OH- ]

= - log (1.37 x 10-13)

= - log 1.37 - log 10-13

= -0.137 + 13

pOH = 12.863

d) Given, [OH−] = 4.32 x 10-4 M

Therefore pOH = - log[OH- ]

= - log (4.32 x 10-4)

= - log 4.32 - log 10-4

= -0.635 + 4

pOH = 3.365

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