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pOH Calculations Using a Calculator
You will need to locate the log_{10} button on your calculator.
This button will probably be labelled log or LOG
For pOH calculations, do NOT use the button labelled ln or LN
On AUSeTUTE's calculator the log button is positioned in the top left hand corner of the calculator.
On the calculator shown below, we have disabled all the buttons except the log_{10} button.
pOH can be positive or negative!
Check the calculations below:
[OH^{}] mol L^{1} 
1 x 10^{3}  5 x 10^{3}  1 x 10^{2}  5 x 10^{2}  1 x 10^{1}  5 x 10^{1}  1 x 10^{0}  5 x 10^{0}  1 x 10^{1} 
(= 0.001)  (= 0.005)  (= 0.01)  (= 0.05)  (= 0.1)  (0.5)  (= 1)  (= 5)  (= 10) 
pOH 
3  2.3  2  1.3  1  0.3  0  0.7  1 
trend 
positive pOH  pOH = 0  negative pOH 
 [OH^{}] < 1 mol L^{1}
pOH > 0 (pOH is positive)
 [OH^{}] = 1 mol L^{1}
pOH = 0
 [OH^{}] > 1 mol L^{1}
pOH < 0 (pOH is negative)
Notice that a tenfold increase in the concentration of hydroxide ions results in a decrease of 1 pOH unit as shown in the tables below:
[OH^{}]=1 x 10^{3} M  x10=  [OH^{}]=1 x 10^{2} M  x10=  [OH^{}]=1 x 10^{1} M  x10=  [OH^{}]=1 x 10^{0} M  x10=  [OH^{}]=1 x 10^{1} M 
pOH = 3   1 =  pOH = 2   1 =  pOH = 1   1 =  pOH = 0   1 =  pOH = 1 
[OH^{}]=5 x 10^{3} M  x10=  [OH^{}]=5 x 10^{2} M  x10=  [OH^{}]=5 x 10^{1} M  x10=  [OH^{}]=5 x 10^{0} M 
pOH = 2.3   1 =  pOH = 1.3   1 =  pOH = 0.3   1 =  pOH = 0.7 
pOH Graphs
Using the formula (equation) pOH = log_{10}[OH^{}] we can calculate the pOH of solutions with varying OH^{} concentrations and plot these values on a graph.
An example is shown below:
Data 
Graph 
Trends 
[OH^{}] mol L^{1} 
log[OH^{}]

=

pOH

0.05 
log[0.05] 
= 
1.30 
0.10 
log[0.10] 
= 
1.00 
0.15 
log[0.15] 
= 
0.82 
0.20 
log[0.20] 
= 
0.70 
0.25 
log[0.25] 
= 
0.60 
0.30 
log[0.30] 
= 
0.52 
0.35 
log[0.35] 
= 
0.46 
0.40 
log[0.40] 
= 
0.40 
0.45 
log[0.45] 
= 
0.35 
0.50 
log[0.50] 
= 
0.30 

pOH  Relationship Between pOH and [OH^{}]
[OH^{}] mol L^{1} 

Increasing [OH^{}] decreases pOH.
Higher [OH^{}] gives a lower pOH.
Decreasing [OH^{}] increases pOH.
Lower [OH^{}] gives a higher pOH.

If you place your mouse over any of the points in the graph, a small box should appear to show you the values of the hydroxide ion concentration and the pOH at that point.
Worked Examples
(based on the StoPGoPS approach to problem solving in chemistry.)
Question 1. A 0.01 mol L^{1} solution of sodium hydroxide has a hydroxide ion concentration of 0.01 mol L^{1}.
Calculate the pOH of this solution.
 What have you been asked to do?
Calculate the pOH
pOH = ?
 What information (data) have you been given?
Extract the data from the question:
hydroxide ion concentration = [OH^{}] = 0.01 mol L^{1}
 What is the relationship between what you know and what you need to find out?
Write the equation (formula) for calculating pOH:
pOH = log_{10}[OH^{}]
 Substitute in the values and solve:
pOH = log_{10}[0.01]
= 2.00
 Is your answer plausible?
Use the value for the concentration of OH^{} to calculate the pOH and compare this value to the value given in the question:
[OH^{}] = 10^{pOH} = 10^{2} = 0.01 mol L^{1}
Since this value is the same as the one given in the question, we are confident our answer is correct.
 State your solution to the problem:
pOH = 2.00
Question 2. 0.25 moles of hydroxide ions are present in 500.0 mL of an aqueous solution of potassium hydroxide.
Calculate the pOH of the solution.
 What have you been asked to do?
Calculate the pOH
pOH = ?
 What information (data) have you been given?
Extract the data from the question:
moles OH^{} = 0.25 mol
volume = 500 mL
 What is the relationship between what you know and what you need to find out?
Calculate the concentration of hydroxide ions in mol L^{1} (molarity):
hydroxide ion concentration = moles OH^{} ÷ volume of solution in litres
moles OH^{} = 0.25 mol
volume of solution = 500.0 mL
Convert volume in mL to L by dividing by 1000:
volume = 500 mL ÷ 1000 mL L^{1} = 0.50 L
hydroxide ion concentration = [OH^{}] = 0.25 mol ÷ 0.50 L = 0.50 mol L^{1}
Write the equation (formula) for calculating pOH:
pOH = log_{10}[OH^{}]
 Substitute in the values and solve:
pOH = log_{10}[0.5]
= 0.30
 Is your answer plausible?
Use the value for the pOH to calculate moles OH^{} and compare this value to the value given in the question:
[OH^{}] = 10^{pOH} = 10^{0.3} = 0.5 mol L^{1}
[OH^{}] = moles ÷ volume
so, moles = [OH^{}] × volume = 0.5 × 500/1000 = 0.25 mol
Since this value is the same as the one given in the question, we are confident our answer is correct.
 State your solution to the problem:
pOH = 0.30
Question 3. Beaker A contains an aqueous sodium hydroxide solution with a hydroxide ion concentration of 0.02 mol L^{1}.
Beaker B contains an aqueous potassium hydroxide solution with a hydroxide ion concentration of 0.04 mol L^{1}.
Which beaker contains the base with the highest pOH?
 What have you been asked to do?
Calculate the pOH
pOH = ?
 What information (data) have you been given?
Extract the data from the question:
Beaker A : [OH^{}] = 0.02 mol L^{1}
Beaker B : [OH^{}] = 0.04 mol L^{1}
 What is the relationship between what you know and what you need to find out?
Write a generalisation for the relationship between pOH and [OH^{}] :
A solution with a lower concentration of hydroxide ions will have the higher pOH.
Place the beakers in order of hydroxide ion concentration from highest to lowest:
0.04 mol L^{1}  >  0.02 mol L^{1} 
Beaker B   Beaker A 
higher [OH^{}]   lower [OH^{}] 
 Compare the relative pOH of each solution in each beaker:
0.04 mol L^{1}  >  0.02 mol L^{1} 
Beaker B   Beaker A 
higher [OH^{}]   lower [OH^{}] 
lower pOH   higher pOH 
Beaker A has a lower concentration of hydroxide ions.
Beaker A will have the higher pOH.
 Is your answer plausible?
beaker A: pOH = log_{10}[OH^{}] = log_{10}[0.02] = 1.7
beaker B: pOH = log_{10}[OH^{}] = log_{10}[0.04] = 1.4
1.7 > 1.4
Beaker A has the higher pOH
Since the answer we got by calculating the pOH of each solution is the same as the answer we got above, we are confident that Beaker A has the higher pOH.
 State your solution to the problem:
Beaker A has the higher pOH