Question

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

pH = 8.76
pH = 11.32
pH = 2.80

Answer 1

We know that pH + pOH = 14 at 25°C.

Using this relation, we can calculate the pOH for each given pH:

For pH = 8.76:

pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 8.76
pOH = 5.24

Now we can use the relation Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C to find [H3O+] and [OH-]:

Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+][10^-pOH]
1.0 x 10^-14 = [H3O+][10^-5.24]
[H3O+] = 10^-8.76
[H3O+] = 1.58 x 10^-9 M

[OH-] = Kw/[H3O+]
[OH-] = 1.0 x 10^-14 / 1.58 x 10^-9
[OH-] = 6.33 x 10^-6 M

Therefore, for a solution with pH 8.76, [H3O+] = 1.58 x 10^-9 M and [OH-] = 6.33 x 10^-6 M.

For pH = 11.32:

pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 11.32
pOH = 2.68

Now we can use the relation Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C to find [H3O+] and [OH-]:

Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+][10^-pOH]
1.0 x 10^-14 = [H3O+][10^-2.68]
[H3O+] = 10^-11.32
[H3O+] = 2.15 x 10^-12 M

[OH-] = Kw/[H3O+]
[OH-] = 1.0 x 10^-14 / 2.15 x 10^-12
[OH-] = 4.65 x 10^-3 M

Therefore, for a solution with pH 11.32, [H3O+] = 2.15 x 10^-12 M and [OH-] = 4.65 x 10^-3 M.

For pH = 2.80:

pH + pOH = 14
pOH = 14 - pH
pOH = 14 - 2.80
pOH = 11.20

Now we can use the relation Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25°C to find [H3O+] and [OH-]:

Kw = [H3O+][OH-]
1.0 x 10^-14 = [H3O+][10^-pOH]
1.0 x 10^-14 = [H3O+][10^-11.20]
[H3O+] = 10^-2.80
[H3O+] = 1.58 x 10^-3 M

[OH-] = Kw/[H3O+]
[OH-] = 1.0 x 10^-14 / 1.58 x 10^-3
[OH-] = 6.33 x 10^-12 M

Therefore, for a solution with pH 2.80, [H3O+] = 1.58 x 10^-3 M and [OH-] = 6.33 x 10^-12 M.

Answer 2

For all three cases, we can use the fact that pH + pOH = 14 to find pOH, and then use pOH to calculate [OH−] and [H3O+].

Case 1: pH = 8.76

pOH = 14 - pH = 14 - 8.76 = 5.24

[OH−] = 10^(-pOH) = 10^(-5.24) = 5.2 × 10^(-6) M

[H3O+] = 1 × 10^(-14) / [OH−] = 1 × 10^(-14) / (5.2 × 10^(-6)) = 1.9 × 10^(-9) M

Therefore, [H3O+] = 1.9 × 10^(-9) M and [OH−] = 5.2 × 10^(-6) M.

Case 2: pH = 11.32

pOH = 14 - pH = 14 - 11.32 = 2.68

[OH−] = 10^(-pOH) = 10^(-2.68) = 2.1 × 10^(-3) M

[H3O+] = 1 × 10^(-14) / [OH−] = 1 × 10^(-14) / (2.1 × 10^(-3)) = 4.8 × 10^(-12) M

Therefore, [H3O+] = 4.8 × 10^(-12) M and [OH−] = 2.1 × 10^(-3) M.

Case 3: pH = 2.80

pOH = 14 - pH = 14 - 2.80 = 11.20

[OH−] = 10^(-pOH) = 10^(-11.20) = 6.3 × 10^(-12) M

[H3O+] = 1 × 10^(-14) / [OH−] = 1 × 10^(-14) / (6.3 × 10^(-12)) = 1.6 × 10^(-3) M

Therefore, [H3O+] = 1.6 × 10^(-3) M and [OH−] = 6.3 × 10^(-12) M.