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+]
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- 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-]
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- 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
- Step 1: Calculate hydrogen concentration by rearranging the equation for pH
