Acids and Bases — IB Chemistry SL SL Study Guide

For: IB Chemistry SL candidates sitting IB Chemistry SL.

Covers: Bronsted-Lowry acid-base definitions, pH scale and ionic product of water ($K_w$), comparisons between strong and weak acids and bases, and step-by-step pH calculations for strong monoprotic acids and bases.

You should already know: IGCSE Chemistry, basic algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the IB Chemistry SL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Acids and Bases?

Acids and bases are a core class of reactive chemical species defined by their ability to transfer or accept hydrogen ions (protons, $H^{+}$) in aqueous solution, forming the basis of common reactions including neutralization, buffer action, and titration. This topic makes up ~11% of your IB Chemistry SL Paper 1 and 2 marks, with frequent 3-6 mark calculation and explanation questions appearing in every exam series.

2. Bronsted-Lowry definition

The Bronsted-Lowry theory is the standard definition of acids and bases used in IB Chemistry, focusing on proton transfer between species:

• A Bronsted-Lowry acid is a proton ($H^{+}$) donor: it releases a hydrogen ion when reacting with a base
• A Bronsted-Lowry base is a proton ($H^{+}$) acceptor: it gains a hydrogen ion when reacting with an acid

Free $H^{+}$ ions do not exist in aqueous solution: they bond to water molecules to form hydronium ions ($H_3{O}^{+}$), which you will often see written instead of $H^{+}$ for technical accuracy. All Bronsted-Lowry acid-base reactions form conjugate acid-base pairs:

• When an acid donates a proton, the remaining species is its conjugate base (it can accept a proton to reform the original acid)
• When a base accepts a proton, the new species is its conjugate acid (it can donate a proton to reform the original base) Water is amphoteric, meaning it can act as both an acid and a base depending on the other reactant.

Worked example

For the reaction between ethanoic acid and ammonia: CH_3COOH_{(aq)} + NH_{3(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + NH_4^+_{(aq)}

1. $C{H}_{3}COOH$ is the acid (donates 1 $H^{+}$), its conjugate base is $C{H}_{3}CO{O}^{-}$
2. $N{H}_3$ is the base (accepts 1 $H^{+}$), its conjugate acid is $N{H}_4^{+}$

Exam tip: Examiners often ask you to label conjugate pairs in given reactions. Always track gain/loss of one hydrogen atom, not just charge change, to avoid mistakes.

3. pH scale, $K_w$

The pH scale is a logarithmic measure of hydrogen ion concentration in solution, designed to avoid working with very small exponential values. The core formula for pH is: $$pH=-{\log }_{10}[H^{+}]$$ where $[H^{+}]$ is the concentration of hydrogen (or hydronium) ions in $mold{m}^{-3}$. The logarithmic scale means a change of 1 pH unit corresponds to a 10x change in $[H^{+}]$. At 298 K (25°C, standard lab temperature):

• pH < 7 = acidic solution (higher $[H^{+}]$ than $[O{H}^{-}]$)
• pH = 7 = neutral solution (equal $[H^{+}]$ and $[O{H}^{-}]$)
• pH > 7 = alkaline solution (lower $[H^{+}]$ than $[O{H}^{-}]$)

The ionic product of water ($K_w$)

Pure water dissociates slightly in a reversible endothermic reaction: $$2H_2{O}_{(l)}\rightleftharpoons H_3{O}_{(aq)}^{+}+O{H}_{(aq)}^{-}$$ Since the concentration of undissociated water is almost constant, we simplify the equilibrium constant to get $K_w$, the ionic product of water: $$K_w=[H^{+}][O{H}^{-}]$$ At 298 K, $K_w=1.00\times 10^{-14}mo{l}^{2}d{m}^{-6}$ — this value must be memorized for the exam, as it is not always provided. $K_w$ increases with temperature, so neutral pH is less than 7 at temperatures above 298 K, a common trick question in Paper 1.

Worked example

Calculate the pH of a solution with $[O{H}^{-}]=2.5\times 10^{-4}mold{m}^{-3}$ at 298 K:

1. Rearrange $K_w$ to find $[H^{+}]$: $[H^{+}]=\frac{K_w}{[O{H}^{-}]}=\frac{1.00\times 10^{-14}}{2.5\times 10^{-4}}=4.0\times 10^{-11}mold{m}^{-3}$
2. Calculate pH: $pH=-{\log }_{10}(4.0\times 10^{-11})=10.40$

4. Strong vs weak acids and bases

The strength of an acid or base refers to its degree of dissociation in aqueous solution, which is unrelated to its concentration (moles of solute per $d{m}^3$ of solution).

• Strong acids/bases: Dissociate completely (100% ionization) in aqueous solution, so no equilibrium is established. For a strong acid HA, the reaction goes to completion: $H{A}_{(aq)}\to H_{(aq)}^{+}+A_{(aq)}^{-}$
• Weak acids/bases: Dissociate only partially in aqueous solution, forming a dynamic equilibrium where most species remain undissociated. For a weak acid HA: $H{A}_{(aq)}\rightleftharpoons H_{(aq)}^{+}+A_{(aq)}^{-}$

Common species to memorize

Experimental distinction (common exam question)

For solutions of equal concentration, you can tell strong and weak species apart using 3 standard tests:

1. pH measurement: Strong acids have lower pH, strong bases have higher pH than their weak counterparts
2. Conductivity: Strong species have higher electrical conductivity, as they produce more free ions in solution
3. Reaction rate: Strong acids react faster with active metals (e.g. Mg) or carbonates, due to higher $[H^{+}]$

Worked example

Equal volumes of 0.1 $mold{m}^{-3}$ HCl (strong acid) and 0.1 $mold{m}^{-3}$ $C{H}_{3}COOH$ (weak acid) are reacted with excess magnesium ribbon. State and explain two differences in observations:

1. The HCl reaction produces bubbles of $H_2$ faster: HCl is fully dissociated, so higher $[H^{+}]$ leads to higher collision frequency with Mg atoms
2. Both reactions produce the same total volume of $H_2$: both acids have the same number of moles of $H^{+}$ available to react, they just release it at different rates

5. pH calculations for strong acids/bases

For IB Chemistry SL, you are only required to calculate pH for strong acids and bases (weak species pH calculations are exclusive to HL). All calculations assume 298 K unless stated otherwise, and you can ignore $H^{+}$ from water dissociation as it is negligible for all standard SL questions.

Strong monoprotic acids

Since dissociation is complete, $[H^{+}]=$ concentration of the acid ($c_{acid}$). The formula simplifies to: $$pH=-{\log }_{10}(c_{acid})$$ For diprotic strong acids like $H_{2}S{O}_4$, SL assumes full dissociation of both $H^{+}$ ions, so $[H^{+}]=2\times c_{acid}$.

Strong bases

Strong bases dissociate to release $O{H}^{-}$ ions, so first calculate $[O{H}^{-}]$:

• For monobasic bases like NaOH: $[O{H}^{-}]=c_{base}$
• For diacidic bases like $Ba(OH{)}_2$: $[O{H}^{-}]=2\times c_{base}$

You can then use one of two methods to find pH:

1. Use $K_w$ to find $[H^{+}]=\frac{K_w}{[O{H}^{-}]}$, then calculate pH directly
2. Calculate pOH first: $pOH=-{\log }_{10}[O{H}^{-}]$, then use $pH+pOH=14$ (valid only at 298 K)

Worked example 1

Calculate the pH of 0.015 $mold{m}^{-3}$ $HN{O}_3$ at 298 K:

1. $HN{O}_3$ is strong monoprotic, so $[H^{+}]=0.015mold{m}^{-3}$
2. $pH=-{\log }_{10}(0.015)=1.82$ (always give pH to 2 decimal places for SL)

Worked example 2

Calculate the pH of 0.001 $mold{m}^{-3}$ $Ba(OH{)}_2$ at 298 K:

1. $Ba(OH{)}_2$ is diacidic, so $[O{H}^{-}]=2\times 0.001=0.002mold{m}^{-3}$
2. $pOH=-{\log }_{10}(0.002)=2.70$
3. $pH=14-2.70=11.30$

Worked example 3 (neutralization calculation)

20 $c{m}^3$ of 0.2 $mold{m}^{-3}$ HCl is mixed with 30 $c{m}^3$ of 0.1 $mold{m}^{-3}$ NaOH. Calculate the pH of the resulting solution:

1. Moles of $H^{+}$ = $0.02d{m}^3\times 0.2mold{m}^{-3}=0.004mol$
2. Moles of $O{H}^{-}$ = $0.03d{m}^3\times 0.1mold{m}^{-3}=0.003mol$
3. Excess $H^{+}$ = $0.004-0.003=0.001mol$, total volume = $0.05d{m}^3$
4. $[H^{+}]=\frac{0.001}{0.05}=0.02mold{m}^{-3}$
5. $pH=-{\log }_{10}(0.02)=1.70$

6. Common Pitfalls (and how to avoid them)

• Wrong move: Confusing conjugate pairs by tracking charge instead of H+ gain/loss. Why: Students assume charge difference is the defining feature of conjugate species, rather than proton transfer. Correct move: Always check which species lost 1 H+ (acid → conjugate base) and which gained 1 H+ (base → conjugate acid); charge change is a side effect, not a definition.
• Wrong move: Assuming neutral pH is always 7, regardless of temperature. Why: Most textbook examples use 298 K, so students memorize pH 7 = neutral without context. Correct move: Neutral = $[H^{+}]=[O{H}^{-}]$, which only equals pH 7 at 298 K. If a different $K_w$ is provided, calculate neutral pH as $-{\log }_{10}(\sqrt{K_w})$.
• Wrong move: Mixing up strength and concentration, using terms like "strong ethanoic acid" to refer to a concentrated solution. Why: Students use everyday definitions of "strong" instead of the chemical definition. Correct move: Never use "strong" and "concentrated" interchangeably; strength refers to dissociation degree, concentration refers to moles of solute per volume.
• Wrong move: Forgetting to multiply ion concentrations by 2 for diprotic acids or diacidic bases. Why: Students default to monoprotic/monobasic assumptions without checking the formula. Correct move: Always count the number of H+ or OH- ions per formula unit of the strong species before calculating ion concentration.
• Wrong move: Rounding pH to 1 decimal place or a whole number, losing accuracy marks. Why: Students treat the pH scale as integer values. Correct move: Always give pH values to 2 decimal places for SL, unless the question explicitly asks for fewer.

7. Practice Questions (IB Chemistry SL Style)

Question 1 (2 marks)

For the reaction below, label the two conjugate acid-base pairs: HNO_{3(aq)} + H_2O_{(l)} \rightleftharpoons NO_3^-_{(aq)} + H_3O^+_{(aq)}

Solution

• Pair 1: Acid = $HN{O}_3$, conjugate base = $N{O}_3^{-}$ (1 mark)
• Pair 2: Base = $H_{2}O$, conjugate acid = $H_3{O}^{+}$ (1 mark)

Question 2 (3 marks)

Calculate the pH of a 0.008 $mold{m}^{-3}$ solution of $Ca(OH{)}_2$ at 298 K, where $K_w=1.00\times 10^{-14}mo{l}^{2}d{m}^{-6}$.

Solution

1. $Ca(OH{)}_2$ is a diacidic strong base, so $[O{H}^{-}]=2\times 0.008=0.016mold{m}^{-3}$ (1 mark)
2. $pOH=-{\log }_{10}(0.016)=1.80$, so $pH=14-1.80=12.20$ (1 mark for correct calculation, 1 mark for 2 decimal place pH)

Question 3 (4 marks)

A student has two unlabelled 0.1 $mold{m}^{-3}$ solutions, one of potassium hydroxide (strong base) and one of ammonia (weak base). Describe two experimental tests they can carry out to distinguish the solutions, and state the expected observations for each test.

Solution

Any two of the following, 2 marks per test:

1. pH test: Use a calibrated pH probe to measure the pH of each solution. KOH will have a higher pH (~13) than ammonia (~11.1), as KOH fully dissociates to give a higher $[O{H}^{-}]$.
2. Conductivity test: Use a conductivity meter to test electrical conductivity of each solution. KOH will show a much higher conductivity reading, as it produces more free ions in solution.
3. Reaction with ammonium chloride: Add 1 $g$ of ammonium chloride to each solution. The ammonia solution will show little to no visible change, while KOH will release pungent ammonia gas, as strong base reacts with ammonium ions to form $N{H}_3$.

8. Quick Reference Cheatsheet

9. What's Next

This topic forms the foundation for all remaining acid-base content in the IB Chemistry SL syllabus, including acid-base titrations, salt solution pH, and acid deposition (a required core environmental chemistry topic). You will apply the pH calculation rules you learned here to solve titration stoichiometry questions, which are common 6-8 mark extended response questions in Paper 2, and you will use the strong vs weak distinction to interpret buffer action and the environmental impacts of acid rain later in the course. A firm grasp of these core concepts will also reduce the amount of new content you need to learn if you decide to move to HL Chemistry later in your course.

If you are stuck on any part of this topic, or want to practice more exam-style questions tailored to your weak areas, you can ask Ollie, our AI tutor, for personalized help any time. You can also find more study guides and past paper practice for all IB Chemistry SL topics on the homepage to help you prepare for your exams and score the highest possible grade.