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pH Calculations Using a Calculator
You will need to locate the log10 button on your calculator.
This button will probably be labelled log or LOG
For pH calculations, do NOT use the button labelled ln or LN
On AUS-e-TUTE'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 log10 button.
pH can be positive or negative!
Check the calculations below:
[H+] mol L-1 |
1 × 10-3 | 5 × 10-3 | 1 × 10-2 | 5 × 10-2 | 1 × 10-1 | 5 × 10-1 | 1 × 100 | 5 × 100 | 1 × 101 |
(= 0.001) | (= 0.005) | (= 0.01) | (= 0.05) | (= 0.1) | (0.5) | (= 1) | (= 5) | (= 10) |
pH |
3 | 2.3 | 2 | 1.3 | 1 | 0.3 | 0 | -0.7 | -1 |
trend |
positive pH | pH = 0 | negative pH |
- [H+] < 1 mol L-1 pH > 0 (pH is positive)
- [H+] = 1 mol L-1 pH = 0
- [H+] > 1 mol L-1 pH < 0 (pH is negative)
Notice that a tenfold increase in the concentration of hydrogen ions results in a decrease of 1 pH unit as shown in the tables below:
[H+]=1 × 10-3 M | ×10= | [H+]=1 × 10-2 M | ×10= | [H+]=1 × 10-1M | ×10= | [H+]=1 × 100 M | ×10= | [H+]=1 × 101 M |
pH = 3 | - 1 = | pH = 2 | - 1 = | pH = 1 | - 1 = | pH = 0 | - 1 = | pH = -1 |
[H+]=5 × 10-3 M | ×10= | [H+]=5 × 10-2 M | ×10= | [H+]=5 × 10-1 M | ×10= | [H+]=5 × 100 M |
pH = 2.3 | - 1 = | pH = 1.3 | - 1 = | pH = 0.3 | - 1 = | pH = -0.7 |
pH Graphs
Using the formula (equation) pH = -log10[H+] we can calculate the pH of solutions with varying H+ concentrations and plot these values on a graph.
An example is shown below:
Data |
Graph |
Trends |
[H+] mol L-1 |
-log[H+] |
= |
pH |
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 |
|
pH | Relationship Between pH and [H+]
[H+] mol L-1 |
|
Increasing [H+] decreases pH.
Higher [H+] gives a lower pH.
Decreasing [H+] increases pH.
Lower [H+] gives a higher pH.
|
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. A 0.01 mol L-1 solution of hydrochloric acid has a hydrogen ion concentration of 0.01 mol L-1.
Calculate the pH of this solution.
- What have you been asked to do?
Calculate pH of solution
pH = ?
- What information (data) have you been given?
Extract the data from the question:
hydrogen ion concentration = [H+] = 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 pH:
pH = -log10[H+]
- Substitute in the values and solve to find pH:
pH = -log10[H+]
= -log10[0.01]
= 2.00
- Is your answer plausible?
Consider how the values of pH change with concentration:
[H+] |
[H+] < 1 M |
[H+] = 1 M |
[H+] > 1 M |
pH |
pH > 0 |
pH = 0 |
pH < 0 |
In the question [H+] = 0.01 M
[H+] < 1 M
therefore pH > 1
So our answer has the right sign (+ rather than -)
Remember that a tenfold change in hydrogen ion concentration results in a change of 1 pH unit:
[H+] = 1 M and pH = 0
[H+] = 1 ÷ 10 = 0.10 M then pH = 0 + 1 = 1
[H+] = 0.10 ÷ 10 = 0.010 M then pH = 1 + 1 = 2
Since the pH value we calculated using the formula (equation) agrees with the logic described above we are confident our solution is correct.
- State your solution to the problem:
pH = 2.00
Question 2. 0.25 moles of hydrogen ions are present in 500.0 mL of an aqueous solution of nitric acid.
Calculate the pH of the solution.
- What have you been asked to do?
Calculate pH of solution
pH = ?
- What information (data) have you been given?
Extract the data from the question:
moles H+ = 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 hydrogen ions in mol L-1 (molarity):
hydrogen ion concentration = moles H+ ÷ volume of solution in litres
moles H+ = 0.25 mol
Convert volume of solution from mL to L by dividing by 1000
volume = 500.0 mL = 500 ÷ 1000 = 0.50 L
hydrogen ion concentration = [H+] = 0.25 mol ÷ 0.50 L = 0.50 mol L-1
Write the equation (formula) for calculating pH:
pH = -log10[H+]
- Substitute in the values and solve to find pH:
pH = -log10[H+]
= -log10[0.50]
= 0.30
- Is your answer plausible?
Consider this:
500 mL is ½ L
So if 500 mL contains 0.25 mol H+, then
2 × 500 mL = 1 L contains 2 × 0.25 = 0.5 mol
therefore [H+] = moles ÷ volume (L) = 0.5 mol ÷ 1 L = 0.5 mol L-1
If [H+] = 0.1 mol L-1 then pH = 1
If [H+] > 0.1 mol L-1 (0.5 > 0.1) then pH < 1 (that is, 0.3 < 1)
(increasing the concentration of H+ decreases the pH)
Since the pH value we calculated using the formula (equation) agrees with the logic described above we are confident our solution is correct.
- State your solution to the problem:
pH = 0.30
Question 3. Beaker A contains sulfuric acid with a hydrogen ion concentration of 0.02 mol L-1.
Beaker B contains hydrochloric acid with a hydrogen ion concentration of 0.04 mol L-1.
Which beaker contains the acid with the lowest pH?
- What have you been asked to do?
Determine which beaker, A or B, has the lowest pH
pH = ?
- What information (data) have you been given?
Extract the data from the question:
Beaker A : [H+] = 0.02 mol L-1
Beaker B : [H+] = 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 pH and [H+] :
A solution with a higher concentration of hydrogen ions will have the lower pH.
Place the beakers in order of hydrogen ion concentration from highest to lowest:
0.04 mol L-1 | > | 0.02 mol L-1 |
Beaker B | | Beaker A |
higher [H+] | | lower [H+] |
Compare the relative pH of each solution in each beaker:
0.04 mol L-1 | > | 0.02 mol L-1 |
Beaker B | | Beaker A |
higher [H+] | | lower [H+] |
lower pH | | higher pH |
- Compare the pH of Beakers A and B:
Beaker B has a higher concentration of hydrogen ions.
Beaker B will have the lower pH.
- Is your answer plausible?
Calculate the pH in each beaker
pH = -log10[H+]
Beaker A: pH = -log10[0.02] = 1.7
Beaker B: pH = -log10[0.04] = 1.4
Beaker B has a lower pH
Since the pH value we calculated using the formula (equation) above agrees with the logic described above to solve the problem we are confident our solution is correct.
- State your solution to the problem:
Solution in Beaker B has the lower pH