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Defining & Using pH & pOH

25oC

Acid

 neutral

Base

  
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH
E
x
a
m
p
l
e
s		 1M HCl 0.1M HCl Gastric juice,
ant venom		 coca cola,
lemon juice,
vinegar		 wine coffee,
toma-
toes		 tap water,
saliva,
cow's milk		 pure water,
NaCl(aq),
KNO3(aq)		 sea water, soap,
baking soda		 deter-
gents,
tooth-
paste		 deter-
gents,
washing soda		 house-
hold cleaner		 0.1M NaOH,
caustic oven cleaner		 1M NaOH +
pOH 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 pOH
  most acidic < ---- ----- ---- --------- least acidic neutral least basic ----- -------- ----- ----- > most basic =14

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) + SO42-(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 (25oC)

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) + SO42-(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

Practice Questions Homework Checker
For AUS-e-TUTE members:
  1. Click on the pH Calculations drill link:
    pH Calculations drill
  2. Enter your username and password if prompted.
  3. Click the "New Question" button to begin the drill.
  4. Worked solutions are provided if you need some help!

Not an AUS-e-TUTE Member?

For AUS-e-TUTE members:

Enter the hydrogen ion
concentration,[H+]		 M
Click Calculate:
pH =
pOH =
[OH-] = M
To start again,
click Reset:
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Related AUS-e-TUTE Topics

Definitions and Properties of Acids and Bases

Acid Dissociation Constants (Ka)

Base Dissociation Constants (Kb)

Ion-Product for Water (Kw)

Acid Base Titration Calculations

Acid Base Titration Curves (graphs)

Indicators for Acid Base Titrations

 
 

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