Please do not block ads on this website.
No ads = no money for us = no free stuff for you!
[H^{+}] 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 pH, 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 [H^{+}] you will need to:
 Enter the number (the pH value) in the textbox
 Multiply the pH by 1 to find pH
 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:
pH 
3  2.3  2  1.3  1  0.3  0  0.7  1 
[H^{+}] mol L^{1} 
1 × 10^{3}  5 × 10^{3}  1 × 10^{2}  5 × 10^{2}  1 × 10^{1}  5 × 10^{1}  1 × 10^{0}  5 × 10^{0}  1 × 10^{1} 
(= 0.001)  (= 0.005)  (= 0.01)  (= 0.05)  (= 0.1)  (0.5)  (= 1)  (= 5)  (= 10) 
trend 
decreasing pH → 
increasing hydrogen ion concentration → 
 Decreasing pH increases [H^{+}]
 Increasing pH decreases [H^{+}]
Notice that a change of 1 pH unit results in a tenfold change in the concentration of hydrogen ions results:
pH = 3   1 =  pH = 2   1 =  pH = 1   1 =  pH = 0   1 =  pH = 1 
[H^{+}]=1 × 10^{3} M  x10=  [H^{+}]=1 × 10^{2} M  x10=  [H^{+}]=1 × 10^{1} M  x10=  [H^{+}]=1 × 10^{0} M  x10=  [H^{+}]=1 × 10^{1} M 
pH = 2.3   1 =  pH = 1.3   1 =  pH = 0.3   1 =  pH = 0.7 
[H^{+}]=5 × 10^{3} M  x10=  [H^{+}]=5 × 10^{2} M  x10=  [H^{+}]=5 × 10^{1} M  x10=  [H^{+}]=5 × 10^{0} M 
pH  [H^{+}] Graphs
Using the formula (equation) [H^{+}] = 10^{pH} we can calculate the [H^{+}] in mol L^{1} of solutions with varying pH values and plot these values on a graph.
An example is shown below:
Data 
Graph 
Trends 
pH 
10^{pH} 
= 
[H^{+}] 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 

[H^{+}] mol L^{1}  Relationship Between [H^{+}] and pH
pH 

Increasing pH decreases [H^{+}].
Higher pH gives a lower [H^{+}].
Decreasing pH increases [H^{+}].
Lower pH gives a higher [H^{+}].

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 hydrogen ion concentration and the pH at that point.
Worked Examples
(based on the StoPGoPS approach to problem solving in chemistry.)
Question 1. The pH of an aqueous solution of an acid is determined to be 5.8
What is the concentration of hydrogen ions in mol L^{1} in this acidic solution?
 What have you been asked to do?
Calculate concentration of hydrogen ions
[H^{+}] = ? mol L^{1}
 What information (data) have you been given?
Extract the data from the question:
pH = 5.8
 What is the relationship between what you know and what you need to find out?
Write the equation (formula) for finding [H^{+}] given the pH:
[H^{+}] = 10^{pH}
 Substitute in the values and solve:
[H^{+}] = 10^{5.8}
= 1.6 × 10^{6} mol L^{1}
 Is your answer plausible?
Use the value for the concentration of H^{+} to calculate the pH and compare this value to the value given in the question:
pH = log_{10}[H^{+}] = log_{10}[1.6 × 10^{6}] = 5.8
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:
[H^{+}] = 1.6 × 10^{6} mol L^{1}
Question 2. The pH of an aqueous solution of hydrochloric acid is 3.2
Calculate the moles of hydrogen ions in 0.20 L of this solution.
 What have you been asked to do?
Calculate moles of hydrogen ions
n(H^{+}) = ? mol
 What information (data) have you been given?
Extract the data from the question:
pH = 3.2
volume = 0.20 L
 What is the relationship between what you know and what you need to find out?
Find the concentration of hydrogen ions:
[H^{+}] = 10^{pH}
= 10^{3.2}
= 6.3 × 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 H^{+}:
molarity = 6.3 × 10^{4} mol L^{1}
volume = 0.20 L
moles H^{+} = molarity (mol L^{1}) × volume (L)
= 6.3 × 10^{4} × 0.20
= 1.3 × 10^{4} mol
 Is your answer plausible?
Use the value for the moles of H^{+} to calculate the pH and compare this value to the value given in the question:
[H^{+}] = moles ÷ volume = 1.3 × 10^{4} mol ÷ 0.20 L = 6.5 × 10^{4} mol L^{1}
pH = log_{10}[H^{+}] = log_{10}[6.5 × 10^{4}] = 3.2
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:
n(H^{+}) = 1.3 × 10^{4} mol
Question 3. An aqueous solution of citric acid has a pH of 6.0
Calculate the number of hydrogen ions present in 100 mL of this solution.
 What have you been asked to do?
Calculate number of hydrogen ions
N(H^{+}) = ?
 What information (data) have you been given?
Extract the data from the question:
pH = 6.0
volume = 100 mL
 What is the relationship between what you know and what you need to find out?
Find the concentration of hydrogen ions:
[H^{+}] = 10^{pH} = 10^{6.0} mol L^{1}
Find the moles of H^{+}:
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 H^{+}:
molarity = 10^{6} mol L^{1}
Convert volume in mL to L by dividing by 1000:
volume = 100 mL = 100 × 10^{3} = 0.10 L
moles H^{+} = 10^{6} × 0.10 = 10^{7} mol
 Calculate moles of H^{+}:
Calculate the number of hydrogen ions:
1 mole = Avogadro's Number of particles = 6.02 × 10^{23} particles
10^{7} mol H^{+} = 10^{7} × 6.02 × 10^{23}
= 6.0 × 10^{16} hydrogen ions
 Is your answer plausible?
Use the value for the number of H^{+} to calculate the pH and compare this value to the value given in the question:
moles(H^{+}) = N(H^{+}) ÷ N_{A} = (6.0 × 10^{16}) ÷ (6 × 10^{23}) = 10^{7} mol
[H^{+}] = moles ÷ volume =10^{7} mol ÷100/1000 = 10^{6} mol L^{1}
pH = log_{10}[H^{+}] = log_{10}[10^{6}] = 6
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:
N(H^{+}) = 6.0 × 10^{16}
Question 4. An aqueous solution of hydrochloric acid contains 0.005 moles of hydrogen ions and has a pH of 4.3.
Calculate the volume of the solution in litres.
 What have you been asked to do?
Calculate volume of solution
V(HCl_{(aq)}) = ? L
 What information (data) have you been given?
Extract the data from the question:
pH = 4.3
moles H^{+} = 0.005 mol
 What is the relationship between what you know and what you need to find out?
Find the concentration of hydrogen ions:
[H^{+}] = 10^{pH}
= 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 H^{+} = 0.005 mol
molarity = 5.0 × 10^{5} mol L^{1}
volume of solution = 0.005/5.0 × 10^{5}
= 100 L
 Is your answer plausible?
Use the value for the volume of HCl_{(aq)} to calculate the pH and compare this value to the value given in the question:
[H^{+}_{(aq)}] = [HCl_{(aq)}] = moles ÷ volume (L) = 0.005 ÷ 100 = 5 × 10^{5}
pH = log_{10}[H^{+}] = log_{10}[5 × 10^{5}] = 4.3
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:
V(HCl_{(aq)}) = 100 L