The pOH Scale (HL) (DP IB Chemistry)
Revision Note
The pOH Scale
pH
The acidity of an aqueous solution depends on the number of H+ ions in the solution
pH is defined as:
pH = -log10 [H+]
Where [H+] is the concentration of H+ ions in mol dm–3
Similarly, the concentration of H+ of a solution can be calculated if the pH is known by rearranging the above equation to:
[H+] = 10-pH
The pH scale is a logarithmic scale with base 10
For example, pH 5 is 10 times more acidic than pH 6
This means that each value is 10 times the value below it
pH values are usually given to 2 decimal places
pOH
The basicity of an aqueous solution depends on the number of hydroxide ions, OH-, in the solution
pOH is defined as:
pOH = -log [OH-]
Where [OH-] is the concentration of hydroxide ions in mol dm–3
Similarly, the concentration of OH- of a solution can be calculated if the pH is known by rearranging the above equation to:
[OH-] = 10-pOH
If you are given the concentration of a basic solution and need to find the pH, this can be done by:
[H+] = Kw / [OH-]
Alternatively, if you are given the [OH-] and calculate the pOH, the pH can be found by:
pH = 14 - pOH
As we can see, pH and pOH are interlinked and at all temperatures, pH + pOH = pKw
Relationship between H+, OH–, pH and pOH
To make a conversion, follow the arrow and equation given, so to convert OH– (aq) to pOH use pOH = -log10[OH–]
Worked Example
pH and H+ calculations
Find the pH when the hydrogen ion concentration is 1.60 x 10-4 mol dm-3
Find the hydrogen ion concentration when the pH is 3.10
Answers:
The pH of the solution is:
pH = -log [H+]
pH = -log 1.6 x 10-4
pH = 3.80
The hydrogen concentration can be calculated by rearranging the equation for pH
pH = -log [H+]
[H+] = 10-pH
[H+] = 10-3.10
[H+] = 7.94 x 10-4 mol dm-3
Worked Example
pH calculations of a strong alkali
Calculate the pH of 0.15 mol dm-3 sodium hydroxide, NaOH
Calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50
Answers:
Sodium hydroxide is a strong base which ionises as follows:
NaOH (aq) → Na+ (aq) + OH– (aq)
The pH of the solution is:
[H+] = Kw ÷ [OH–]
[H+] = (1 x 10-14) ÷ 0.15 = 6.66 x 10-14
pH = -log 6.66 x 10-14 = 13.17pH = -log[H+]
To calculate the hydroxide concentration of a solution of sodium hydroxide when the pH is 10.50:
Step 1: Calculate hydrogen concentration by rearranging the equation for pH
pH = -log[H+]
[H+] = 10-10.50
[H+] = 3.16 x 10-11 mol dm-3[H+] = 10-pH
Step 2: Rearrange the ionic product of water to find the concentration of hydroxide ions
Kw = [H+] [OH–]
[OH–] = Kw ÷ [H+]
Step 3: Substitute the values into the expression to find the concentration of hydroxide ions
Since Kw is 1.00 x 10-14
[OH–] = (1 x 10-14) ÷ (3.16 x 10-11)
[OH–] = 3.16 x 10-4 mol dm-3
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