Difficulty: Intermediate | Prerequisites: Strong vs weak acids and bases, ionic dissociation, molarity, stoichiometry.
The Brønsted-Lowry model redefines acids and bases in terms of proton (H+) transfer rather than the production of H+ or OH- in water. It is broader than the Arrhenius model and covers reactions that happen in non-aqueous solvents too. Titration calculations, which close out this unit, apply acid-base stoichiometry to determine unknown concentrations. Together these topics form the core of what exams test in Unit 4's second half.
A Brønsted-Lowry acid donates a proton; a Brønsted-Lowry base accepts one. Every acid-base reaction produces a conjugate pair. Some species (like water, HCO3- and H2PO4-) can act as either acid or base depending on the partner. Titration problems use the moles-of-acid-equal-moles-of-base relationship (adjusted for stoichiometry) to find unknown concentrations.
Brønsted-Lowry acid
A proton (H+) donor. In simple terms, it is any species that gives away a hydrogen ion to another species.
Brønsted-Lowry base
A proton (H+) acceptor. It is any species that takes a hydrogen ion from another species.
Conjugate acid-base pair
Two species that differ by exactly one proton. When an acid donates H+, what remains is its conjugate base. When a base accepts H+, what forms is its conjugate acid.
Amphoteric / amphiprotic species
A substance that can act as either an acid or a base depending on the reaction partner. Water is the classic example: it donates H+ to strong bases and accepts H+ from acids.
Titration
A laboratory technique in which a solution of known concentration (the titrant) is added to a solution of unknown concentration until the reaction is complete (the endpoint). The volumes and concentrations are then used to calculate the unknown.
Endpoint (equivalence point)
The point in a titration at which the moles of acid exactly equal the moles of base (accounting for stoichiometry). In simple terms, this is when the reaction is perfectly finished, with neither reactant in excess.
Molarity (M)
Concentration expressed as moles of solute per litre of solution. This is the unit you will use in every titration calculation.
Look at what happens to each species in the equation. The one that loses an H+ is the acid; the one that gains an H+ is the base.
Example: HCN(aq) + H2O(l) ⇌ CN-(aq) + H3O+(aq). HCN donates a proton to water, so HCN is the acid and H2O is the base.
A species that has no removable H+ can never act as a Brønsted acid (e.g. Na3PO4 in solution: the Na+ and PO4 3- ions are present, and PO4 3- has no H to donate).
A species whose ions cannot accept a proton can never act as a Brønsted base (e.g. NH4Cl: the NH4+ ion is an acid, and Cl- is the conjugate base of a strong acid and is far too weak a base to accept a proton under normal conditions).
Every Brønsted acid-base reaction involves two conjugate pairs. To find the conjugate base of a species, remove one H+ and reduce the charge by 1. To find the conjugate acid, add one H+ and increase the charge by 1.
Conjugate base of H2O: OH-
Conjugate acid of HPO4 2-: H2PO4-
Conjugate base of [Fe(H2O)6]3+: [Fe(H2O)5OH]2+ (one coordinated water loses H+)
Species like HCO3-, H2PO4- and HPO4 2- can donate or accept a proton.
HCO3- acting as an acid: HCO3-(aq) + H2O(l) ⇌ CO3 2-(aq) + H3O+(aq)
H2PO4- acting as a base: H2PO4-(aq) + H2O(l) ⇌ H3PO4(aq) + OH-(aq)
HPO4 2- acting as a base: HPO4 2-(aq) + H2O(l) ⇌ H2PO4-(aq) + OH-(aq)
The direction depends on the reaction partner. With a stronger acid, the amphiprotic species acts as a base; with a stronger base, it acts as an acid.
At the endpoint, moles of acid = moles of base (for a 1 : 1 reaction).
Macid x Vacid = Mbase x Vbase
Solve for the unknown. Keep volumes in the same units (mL is fine as long as both are mL).
Example: 17.5 mL of CH3CO2H requires 29.6 mL of 0.250 M NaOH. Moles NaOH = 0.0296 x 0.250 = 0.00740 mol. Since the ratio is 1 : 1, moles acetic acid = 0.00740. Macid = 0.00740 / 0.0175 = 0.423 M.
For a diprotic acid like H2SO4, each mole of acid provides 2 moles of H+.
Macid x Vacid x 2 = Mbase x Vbase
Example: 31.5 mL of H2SO4 titrated with 23.9 mL of 0.0134 M NaOH. Moles NaOH = 0.0239 x 0.0134 = 0.000320 mol. Moles H2SO4 = 0.000320 / 2 = 0.000160 mol. Macid = 0.000160 / 0.0315 = 0.00508 M.
Monoprotic acid-base titration (1 : 1 stoichiometry):
M_{\text{acid}} \times V_{\text{acid}} = M_{\text{base}} \times V_{\text{base}}Diprotic acid titrated with a monoprotic base:
M_{\text{acid}} \times V_{\text{acid}} \times 2 = M_{\text{base}} \times V_{\text{base}}General form for any stoichiometry:
n_{\text{acid}} \times M_{\text{acid}} \times V_{\text{acid}} = n_{\text{base}} \times M_{\text{base}} \times V_{\text{base}}where n = number of moles of H+ (or OH-) per formula unit.
Titrations are used routinely in food science to measure the acidity of juices and wines, in environmental labs to test water quality, and in pharmaceutical quality control to verify the concentration of active ingredients. The Brønsted-Lowry framework underpins how buffer solutions work in your bloodstream, keeping pH tightly regulated around 7.4.
Students confuse "acting as a base" with "being a base." An amphiprotic species like HPO4 2- acting as a base means it accepts a proton in that particular reaction; it does not mean it is always a base.
When asked for the conjugate base, students sometimes remove an entire OH group instead of a single H+. Conjugate base = remove one H+ and subtract 1 from the charge, nothing more.
For titration of a diprotic acid (H2SO4), students often forget to account for the factor of 2. Each mole of H2SO4 provides 2 moles of H+, so the moles of NaOH needed are double the moles of acid.
Students sometimes assume that the net ionic equation for every acid-base reaction is H+(aq) + OH-(aq) → H2O(l). That is only true when both the acid and the base are strong. If the acid is weak (HCO2H, CH3CO2H, HCN), it stays as a molecule in the net ionic equation.
⚠️ Conjugate pair identification is tested heavily. Be able to look at a reaction and name both pairs.
⚠️ Know which common species are amphiprotic (HCO3-, H2PO4-, HPO4 2-, HSO4-, H2O) and be ready to write equations showing them acting as acid or base.
⚠️ Titration calculations will appear as straightforward plug-and-solve problems. Watch for the diprotic acid twist (H2SO4, H2CO3): the stoichiometric factor changes.
⚠️ The distinction between strong and weak acids matters for writing net ionic equations. If it is a weak acid, keep it as a molecule.
True or false: a Brønsted-Lowry base is a proton donor. (False; a base is a proton acceptor.)
Fill in the blank: the conjugate base of HCO3- is ______. (CO3 2-.)
True or false: Na3PO4 can act as a Brønsted acid in aqueous solution. (False; PO4 3- has no proton to donate.)
Fill in the blank: at the endpoint of a monoprotic titration, moles of acid ______ moles of base. (Equal.)
True or false: for H2SO4 titrated with NaOH, you need twice as many moles of NaOH as moles of H2SO4. (True.)
Q: Which equation depicts HCN behaving as a Brønsted-Lowry acid in water?
A: HCN(aq) + H2O(l) ⇌ CN-(aq) + H3O+(aq). HCN donates a proton to water.
Q: Which equation shows H2PO4- behaving as a Brønsted-Lowry base in water?
A: H2PO4-(aq) + H2O(l) ⇌ H3PO4(aq) + OH-(aq). H2PO4- accepts a proton from water.
Q: In the reaction HS-(aq) + H2O(l) ⇌ S2-(aq) + H3O+(aq), identify the acid and its conjugate base.
A: HS- is the acid; S2- is its conjugate base. HS- donates a proton to water.
Q: What is the conjugate base of [Fe(H2O)6]3+?
A: [Fe(H2O)5OH]2+. One coordinated water molecule donates a proton.
Q: 17.5 mL of acetic acid requires 29.6 mL of 0.250 M NaOH to reach the endpoint. What is the concentration of the acid?
A: Moles NaOH = 0.0296 x 0.250 = 0.00740 mol = moles acid (1 : 1). M = 0.00740 / 0.0175 = 0.423 M.
Q: 25.5 mL of HCl is titrated with 51.2 mL of 0.113 M NaOH. What is the HCl concentration?
A: Moles NaOH = 0.0512 x 0.113 = 0.005786 mol = moles HCl. M = 0.005786 / 0.0255 = 0.227 M.
Q: 31.5 mL of HNO3 is titrated with 23.9 mL of 0.0134 M NaOH. What is the acid concentration?
A: Moles NaOH = 0.0239 x 0.0134 = 0.000320 mol = moles HNO3 (1 : 1). M = 0.000320 / 0.0315 = 0.0102 M.
Q: 31.5 mL of H2SO4 is titrated with 23.9 mL of 0.0134 M NaOH. What is the acid concentration?
A: Moles NaOH = 0.000320 mol. H2SO4 is diprotic, so moles acid = 0.000320 / 2 = 0.000160. M = 0.000160 / 0.0315 = 0.00508 M.
The Brønsted-Lowry model extends the Arrhenius definition from earlier in the course and is a stepping stone to the Lewis acid-base model you will encounter later. Conjugate pairs come back in buffer chemistry, where the weak acid and its conjugate base work together to resist pH changes.
Titration calculations are a direct application of solution stoichiometry from the earlier part of this unit and connect forward to equilibrium and Kw problems.
Brønsted-Lowry acid, Brønsted-Lowry base, proton donor, proton acceptor, conjugate acid, conjugate base, conjugate pair, amphiprotic, amphoteric, water autoionisation, hydronium, hydroxide, titration, endpoint, equivalence point, molarity, monoprotic, diprotic, H2SO4, HCl, HNO3, acetic acid, formic acid, weak acid, strong acid, buffer, net ionic equation, chemistry Unit 4, NYU general chemistry