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[OH^{}] Calculations Using a Calculator
You will need to locate the 10^{x} button on your calculator.
This button will probably be labelled 10^{x} and be above the button labelled log or LOG
For calculations involving pOH, do NOT use the button labelled e^{x} (above the button labelled ln or LN)
On AUSeTUTE's calculator the 10^{x} button is above the log button positioned in the top left hand corner of the calculator.
To use the 10^{x} button to calculate [OH^{}] you will need to:

 Multiply the pOH by 1 to find pOH
 Click the INV button
 Click the log button
On the calculator shown below, we have disabled all the buttons except the INV and log button.
Check the calculations below:
pOH 
3  2.3  2  1.3  1  0.3  0  0.7  1 
[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) 
trend 
decreasing pOH → 
increasing hydroxide ion concentration → 
Note the following relationships between the concentration of hydroxide ions and the pH:
 Decreasing pOH increases [OH^{}]
 Increasing pOH decreases [OH^{}]
Notice that a change of 1 pOH unit results in a tenfold change in the concentration of hydroxide ions results:
pOH = 3   1 =  pOH = 2   1 =  pOH = 1   1 =  pOH = 0   1 =  pOH = 1 
[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 = 2.3   1 =  pOH = 1.3   1 =  pOH = 0.3   1 =  pOH = 0.7 
[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 
[OH^{}] Graphs
Using the formula (equation) [OH^{}] = 10^{pOH} we can calculate the [OH^{}] in mol L^{1} of solutions with varying pOH values and plot these values on a graph.
An example is shown below:
Data 
Graph 
Trends 
pOH 
10^{pOH} 
= 
[OH^{}] mol L^{1} 
0.00 
10^{0.00} 
= 
1.00 
0.05 
10^{0.05} 
= 
0.89 
0.10 
10^{0.10} 
= 
0.79 
0.15 
10^{0.15} 
= 
0.71 
0.20 
10^{0.20} 
= 
0.63 
0.25 
10^{0.25} 
= 
0.56 
0.30 
10^{0.30} 
= 
0.50 
0.35 
10^{0.35} 
= 
0.45 
0.40 
10^{0.40} 
= 
0.40 
0.45 
10^{0.45} 
= 
0.35 

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

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

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. The pOH of an aqueous solution of a base is determined to be 4.4
What is the concentration of hydroxide ions in mol L^{1} in this basic solution?
 What have you been asked to do?
Calculate concentration of hydroxide ions in mol L^{1}
[OH^{}_{(aq)}] = ? mol L^{1}
 What information (data) have you been given?
Extract the data from the question:
pOH = 4.4
 What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding [OH^{}] given the pOH:
[OH^{}] = 10^{pOH}
 Substitute in the values and solve:
[OH^{}] = 10^{pOH}
[OH^{}] = 10^{4.4} = 3.98 × 10^{5} mol L^{1}
 Is your answer plausible?
Use your calculated value of [OH^{}] to find pOH and compare your value to that given in the question:
pOH = log_{10}[OH^{}] = log_{10}[3.98 × 10^{5}] = 4.4
Since this value is the same as that given in the question, we are confident our answer is correct.
 State your solution to the problem:
[OH^{}_{(aq)}] = 3.98 × 10^{5} mol L^{1}
Question 2. The pOH of an aqueous solution of sodium hydroxide is 3.5
Calculate the moles of hydroxide ions in 0.25 L of this solution.
 What have you been asked to do?
Calculate moles of hydroxide ions
n(OH^{}_{(aq)}) = ? mol
 What information (data) have you been given?
Extract the data from the question:
pH = 3.5
volume = 0.25 L
 What is the relationship between what you know and what you need to find out?
Find the concentration of hydroxide ions:
[OH^{}] = 10^{pOH}
= 10^{3.5}
= 3.16 × 10^{4} mol L^{1}
Write the equation (formula) that relates moles, volume and concentration (molarity):
molarity = moles ÷ volume (L)
Rearrange this equation (formula) to find moles:
moles = molarity (mol L^{1}) × volume (L)
 Calculate moles of OH^{}:
molarity = 3.16 × 10^{4} mol L^{1}
volume = 0.25 L
moles OH^{} = 3.16 × 10^{4} × 0.25
= 7.9 × 10^{5} mol
 Is your answer plausible?
Use your calculated value of moles of OH^{} to find pOH and compare your value to that given in the question:
[OH^{}] = moles ÷ volume = (7.9 × 10^{5} ) ÷ 0.25 = 3.16 × 10^{4} mol L^{1}
pOH = log_{10}[OH^{}] = log_{10}[3.16 × 10^{4}] = 3.5
Since this value is the same as that given in the question, we are confident our answer is correct.
 State your solution to the problem:
n(OH^{}_{(aq)}) = 7.9 × 10^{5} mol
Question 3. An aqueous solution of ammonium hydroxide has a pOH of 6.0
Calculate the number of hydroxide ions present in 100 mL of this solution.
 What have you been asked to do?
Calculate number of hydroxide ions
N(OH^{}_{(aq)}) = ?
 What information (data) have you been given?
Extract the data from the question:
pOH = 6.0
volume = 100 mL
 What is the relationship between what you know and what you need to find out?
Find the concentration of hydroxide ions:
[OH^{}] = 10^{pOH} = 10^{6.0} mol L^{1}
Find the moles of OH^{}:
Write the equation (formula) that relates moles, volume and concentration (molarity):
molarity = moles ÷ volume (L)
Rearrange this equation (formula) to find moles:
moles = molarity (mol L^{1}) × volume (L)
Calculate moles of OH^{}:
molarity = 10^{6} mol L^{1}
volume = 100 mL
Convert volume in mL to L by dividing by 1000:
volume = 100 mL ÷ 1000 mLL^{1} = 0.10 L
moles OH^{} = 10^{6} × 0.10 = 10^{7} mol
 Calculate the number of hydroxide ions:
1 mole = Avogadro's Number of particles = 6.02 x 10^{23} particles
10^{7} mol OH^{} = 10^{7} × ( 6.02 × 10^{23})
= 6.0 × 10^{16} hydroxide ions
 Is your answer plausible?
Use your calculated value of number of OH^{} particles to find pOH and compare your value to that given in the question:
moles = N ÷ N_{A} = (6.0 × 10^{16}) ÷ ( 6.02 × 10^{23}) = 10^{7} mol
[OH^{}] = moles ÷ volume in litres = 10^{7} ÷ 100mL/1000mL/L = 10^{6} mol L^{1}
pOH = log_{10}[OH^{}] = log_{10}[10^{6}] = 6
Since this value is the same as that given in the question, we are confident our answer is correct.
 State your solution to the problem:
N(OH^{}_{(aq)}) = 6.0 × 10^{16}
Question 4. An aqueous solution of potassium hydroxide contains 0.0005 moles of hydroxide ions and has a pOH of 4.3.
Calculate the volume of the solution in litres.
 What have you been asked to do?
Calculate volume of the solution in litres
V = ? L
 What information (data) have you been given?
Extract the data from the question:
pOH = 4.3
moles OH^{} = 0.0005 mol
 What is the relationship between what you know and what you need to find out?
Find the concentration of hydroxide ions:
[OH^{}] = 10^{pOH}
= 10^{4.3}
= 5.0 × 10^{5} mol L^{1}
Write the equation (formula) that relates moles, volume and concentration (molarity):
molarity = moles ÷ volume (L)
Rearrange this equation (formula) to find volume:
volume= moles ÷ molarity (mol L^{1})
 Calculate volume of solution:
moles OH^{} = 0.0005 mol
molarity = 5.0 × 10^{5} mol L^{1}
volume of solution = 0.0005 ÷ (5.0 × 10^{5})
= 10 L
 Is your answer plausible?
Use your calculated value of volume of solution to find pOH and compare your value to that given in the question:
[OH^{}] = moles ÷ volume in litres = 0.0005 ÷ 10 = 5 × 10^{5} mol L^{1}
pOH = log_{10}[OH^{}] = log_{10}[5 × 10^{5}] = 4.3
Since this value is the same as that given in the question, we are confident our answer is correct.
 State your solution to the problem:
V = 10 L