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Defining & Using pH & pOH

 

Acid

 neutral

Base

  
pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH
Examples 1M HCl 0.1M HCl Gastric juice
ant venom		 coca cola
lemon juice
vinegar		 wine coffee
tomatoes		 tap water
saliva
cow's milk		 pure water
NaCl(aq)
KNO3(aq)		 sea water soap
baking soda		 detergents
toothpaste		 detergents
washing soda		 household 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-]
    pH = -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) + 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

To calculate the pH of a strong acid
Enter the hydrogen ion concentration, [H+]     The answer will appear below:
Then click Calculate   Calculation :  
To perform another calculation, Click Reset   pH = 

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