6.1.4Additional lessons
The pH Scale — IB Chemistry SL Revision Notes & Topical Questions
5 min read 6 parts Lesson 80 of 98
Curriculum Equation Key pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: pH + pOH = 14
Part 1 of 6Defining pH
The pH scale compresses the huge range of hydrogen ion concentrations into a convenient scale.
Definition
- pH = −log₁₀[H⁺], where [H⁺] is the hydrogen ion concentration in mol dm⁻³.
Inverse relationship
- [H⁺] = 10⁻ᵖᴴ, so a high hydrogen ion concentration gives a low pH.
Logarithmic scale
- A change of one pH unit means [H⁺] changes by a factor of 10, and two units means a factor of 100.
Range
- At 298 K, acidic solutions have pH below 7, neutral solutions have pH 7 and alkaline solutions have pH above 7.
| [H⁺] / mol dm⁻³ | pH |
|---|---|
| 1 × 10⁻¹ | 1 |
| 1 × 10⁻³ | 3 |
| 1 × 10⁻⁷ | 7 |
| 1 × 10⁻¹¹ | 11 |
Part 2 of 6Calculating pH
1
To find pH from concentration, enter the concentration, press log and change the sign.
2
To find [H⁺] from pH, calculate 10 raised to the power of minus pH using the inverse log key.
3
For a strong monoprotic acid, [H⁺] equals the acid concentration.
4
Dilution by a factor of 10 raises the pH of a strong acid by 1 unit.
5
Give pH to the same number of decimal places as significant figures in the concentration.
Part 3 of 6Measuring pH
Measuring pH
pH meter
- A glass electrode gives a digital pH reading and is accurate if calibrated with buffer solutions.
Universal indicator
- A mixture of dyes whose colour changes across the range, giving an approximate pH by comparison with a colour chart.
Limitations
- Indicators are less precise and are hard to use with coloured solutions.
Part 4 of 6Strong Acids, Strong Bases and Dilution
Strong Acids, Strong Bases and Dilution
Strong acid
- For 0.010 mol dm⁻³ HCl, [H⁺] = 0.010 mol dm⁻³ and pH = 2.00.
Tenfold dilution
- Diluting a strong acid by a factor of ten increases its pH by exactly one unit.
Neutral water
- Pure water at 298 K has [H⁺] = 1.0 × 10⁻⁷ mol dm⁻³, which gives pH 7.
Very dilute acid
- As the concentration approaches 10⁻⁷ mol dm⁻³, ionisation of water becomes significant, so the pH never goes above 7.
Part 5 of 6Worked Example
Question Calculate (a) the pH of 0.0050 mol dm⁻³ HCl and (b) [H⁺] in a solution of pH 9.20.
- (a) HCl is a strong acid, so [H⁺] = 0.0050 mol dm⁻³.
- pH = −log(0.0050) = 2.30.
- (b) [H⁺] = 10⁻⁹·²⁰ = 6.3 × 10⁻¹⁰ mol dm⁻³.
Answer
- (a) pH = 2.30.
- (b) [H⁺] = 6.3 × 10⁻¹⁰ mol dm⁻³.
Part 6 of 6Exam Checklist & Tips
6.1.4 – Key Points for the Exam
- pH The negative logarithm to base 10 of the hydrogen ion concentration.
- Strong monoprotic acid An acid that fully ionises and gives one H⁺ per formula unit, such as HCl.
- Universal indicator A mixture of indicators that changes colour over a wide pH range.
- Logarithmic scale A scale in which each unit corresponds to a tenfold change.
Common Mistakes
- Forgetting the minus sign, which gives a negative pH for dilute acids.
- Treating a change of 1 pH unit as a small change; it is a tenfold change in [H⁺].
- Assuming that pH cannot be below 0 or above 14; concentrated solutions can go outside this range.
- Quoting pH to too many decimal places compared with the data given.
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