Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81

1. For pH=8.74;

[H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

[OH⁻] = 5.5 x 10⁻⁶mol dm³

2. For pH=11.38;

[H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

[OH⁻] =2.4 x 10⁻³mol dm³

3. For pH=2.81;

[H₃O⁺]  = 1.55 x 10⁻³ mol dm³

[OH⁻] =6.46 x 10⁻¹²mol dm³

It is given that the values of pH are as follows:

pH  = 8.74

pH = 11.38

pH = 2.81

pH is a measure of the relative amount of free hydrogen and hydroxyl ions in the water.

pH formula:

pH = -log[] or pH = -log[]

We know that, pH + pOH = 14

So, we need to calculate concentration of [H₃O⁺] and [OH⁻] for each of the following pH.

1. For pH=8.74

since  pH = -log[H₃O⁺]

8.74 =  -log[H₃O⁺]

[H₃O⁺] =

[H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

pH + pOH = 14

pOH = 14 - 8.74

pOH = 5.26

pOH = -log[OH⁻]

5.26  = -log[OH⁻]

[OH⁻] =

[OH⁻] = 5.5 x 10⁻⁶mol dm³

2. For pH=11.38

since  pH = -log[H₃O⁺]

11.38 =  -log[H₃O⁺]

[H₃O⁺] =

[H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

pH + pOH = 14

pOH = 14 - 11.38

pOH = 2.62

pOH = -log[OH⁻]

2.62  = -log[OH⁻]

[OH⁻] =

[OH⁻] =2.4 x 10⁻³mol dm³

3. pH=2.81

since  pH = -log[H₃O⁺]

2.81 =  -log[H₃O⁺]

[H₃O⁺] =

[H₃O⁺]  = 1.55 x 10⁻³ mol dm³

pH + pOH = 14

pOH = 14 - 2.81

pOH = 11.19

pOH = -log[OH⁻]

11.19  = -log[OH⁻]

[OH⁻] =

[OH⁻] =6.46 x 10⁻¹²mol dm³

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Explanation:

Given parameters;

pH  = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

      pH = -log[H₃O⁺]

      pOH = -log[OH⁻]

 pH + pOH = 14

Now let us solve;

   pH = 8.74

             since  pH = -log[H₃O⁺]

                           8.74 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻

                             [H₃O⁺]  = 1.82 x 10⁻⁹mol dm³

       pH + pOH = 14

                 pOH = 14 - 8.74

                  pOH = 5.26

                  pOH = -log[OH⁻]

                     5.26  = -log[OH⁻]

                     [OH⁻] = 10

                      [OH⁻] = 5.5 x 10⁻⁶mol dm³

2.  pH = 11.38

             since  pH = -log[H₃O⁺]

                           11.38 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻

                             [H₃O⁺]  = 4.17 x 10⁻¹² mol dm³

           pH + pOH = 14

                 pOH = 14 - 11.38

                  pOH = 2.62

                  pOH = -log[OH⁻]

                     2.62  = -log[OH⁻]

                     [OH⁻] = 10

                      [OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

             since  pH = -log[H₃O⁺]

                           2.81 =  -log[H₃O⁺]

                           [H₃O⁺] = 10⁻

                             [H₃O⁺]  = 1.55 x 10⁻³ mol dm³

           pH + pOH = 14

                 pOH = 14 - 2.81

                  pOH = 11.19

                  pOH = -log[OH⁻]

                     11.19  = -log[OH⁻]

                     [OH⁻] = 10

                      [OH⁻] =6.46 x 10⁻¹²mol dm³

1 a

2 d

3 a

4 c

explanation: