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11 years ago

Hi

# 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

## 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

Thanks

Anurag Kishore