pH

pOH

pH is a measure of the hydrogen ion concentration, [H⁺]

pOH is a measure of the hydroxide ion concentration, [OH^-]

pH is calculated using the following formula:
pH = -log₁₀[H⁺]

pOH is calculated using the following formula:
pOH = -log₁₀[OH^-]

Example 1:

Find the pH of a 0.2mol L^-1
(0.2M) solution of HCl

• Write the balanced equation for the dissociation of the acid
  HCl -----> H⁺(aq) + Cl^-(aq)
• Use the equation to find the [H⁺]:
  0.2 mol L^- HCl produces 0.2 mol L^-1 H⁺ since HCl is a strong acid that fully dissociates
• Calculate pH: pH = -log₁₀[H⁺]
  pH = -log₁₀[0.2] = 0.7

Example 1:

Find the pOH of a 0.1mol L^-
(0.1M) solution of NaOH

• Write the balanced equation for the dissociation of the alkali
  NaOH -----> OH^-(aq) + Na⁺(aq)
• Use the equation to find the [OH^-]:
  0.1 mol L^-1 NaOH produces 0.1 mol L^-1 OH^- since NaOH is a strong alkali that fully dissociates
• Calculate pOH: pOH = -log₁₀[OH^-]
  pOH = -log₁₀[0.1] = 1

Example 2:

Find the pH of a 0.2 mol L^-1
(0.2M) solution of H₂SO₄

• Write the balanced equation for the dissociation of the acid
  H₂SO₄ -----> 2H⁺(aq) + SO₄^2-(aq)
• Use the equation to find the [H⁺]:
  0.2 mol L^-1 H₂SO₄ produces 2 x 0.2 = 0.4 mol L^-1 H⁺ since H₂SO₄ is a strong acid that fully dissociates
• Calculate pH: pH = -log₁₀[H⁺]
  pH = -log₁₀[0.4] = 0.4

Example 2:

Find the pOH of a 0.1mol L^-1
(0.1M) solution of Ba(OH)₂

• Write the balanced equation for the dissociation of the alkali:
  Ba(OH)₂ -----> 2OH^-(aq) + Ba²⁺(aq)
• Use the equation to find the [OH^-]:
  0.1mol L^-1 Ba(OH)₂ produces 2 x 0.1 = 0.2 mol L^-1 OH^- since Ba(OH)₂ is a strong alkali that fully dissociates
• Calculate pOH: pOH = -log₁₀[OH^-]
  pOH = -log₁₀[0.2] = 0.7

Hydrogen ion concentration, [H⁺], can be calculated using the following formula:
[H⁺] = 10^-pH

Hydroxide ion concentration, [OH^-], can be calculated using the following formula:
[OH^-] = 10^-pOH

Example:

Find the [H⁺] of a nitric acid solution with a pH of 3.0
pH= 3.0
[H⁺] = 10^-pH
[H⁺] = 10^-3.0 = 0.001mol L^-1

You can check this answer by using the calculated value [H⁺] in the equation for pH to make sure you arrive at the original pH
pH = -log₁₀[H⁺]
pH = -log₁₀[0.001] = 3
We get the same value for pH using the calculated value for [H⁺], so the calculated value for [H⁺] is correct.

Example:

Find the [OH^-] of a sodium hydroxide solution with a pOH of 1
pOH = 1
[OH^-] = 10^-pOH
[OH^-] = 10^-1 = 0.1 mol L^-1

You can check this answer by using the calculated value [OH^-] in the equation for pOH to make sure you arrive at the original pOH
pOH = -log₁₀[OH^-]
pOH = -log₁₀[0.1] = 1
We get the same value for pOH using the calculated value for [OH^-], so the calculated value for [OH^-] is correct.

pH + pOH = 14

Example A(1):

Find the pH of a solution of sodium hydroxide
that has a pOH of 2

pH = 14 - pOH
pH = 14 - 2 = 12

Example B(1):

Find the pOH of a solution of hydrochloric acid
that has a pH of 3.4

pOH = 14 - pH
pOH = 14 - 3.4 = 10.6

Example A(2):

Find the [H⁺] in a solution of sodium hydroxide
that has a pOH of 1

• Calculate the pH
  pH = 14 - pOH
  pH = 14 - 1 = 13
• Calculate [H⁺]
  [H⁺] = 10^-pH
  [H⁺] = 10^-13 = 10^-13mol L^-1

Example B(2):

Find the [OH^-] of a sulfuric acid solution
with a pH of 3

• Calculate the pOH
  pOH = 14 - pH
  pOH = 14 - 3 = 11
• Calculate [OH^-]
  [OH^-] = 10^-pOH
  [OH^-] = 10^-11 = 10^-11 mol L^-1

Example A(3):

Find the pH of 0.2mol L^-1 sodium hydroxide

• Write the equation for the dissociation of NaOH:
  NaOH -----> Na⁺(aq) + OH^-(aq)
• Use the equation to find [OH^-]:
  0.2mol L^-1 NaOH produces 0.2mol L^-1 OH^- since NaOH is a strong base that fully dissociates
• Calculate the pOH:
  pOH = -log₁₀[OH^-]
  pOH = -log₁₀[0.2] = 0.7
• Calculate pH:
  pH = 14 - pOH
  pH = 14 - 0.7 = 13.3

Example B(3):

Find the pOH of 0.2mol L^-1 sulfuric acid

• Write the equation for the dissociation of H₂SO₄:
  H₂SO₄ -----> 2H⁺(aq) + SO₄^2-(aq)
• Use the equation to find [H⁺]:
  0.2 mol L^-1 H₂SO₄ produces 2 x 0.2 = 0.4 mol L^-1 H⁺ since H₂SO₄ is a strong acid that fully dissociates
• Calculate the pH:
  pH = -log₁₀[H⁺]:
  pH = -log₁₀[0.4] = 0.4
• Calculate pOH:
  pOH = 14 - pH
  pOH = 14 - 0.4 = 13.6