1. The Avogadro Constant and Particle Counting
The mole bridges the atomic scale and laboratory quantities. One mole of any substance contains exactly \(6.022 \times 10^{23}\) particles:
The number of defined entities (atoms, molecules, ions, or electrons) in exactly one mole of a substance, defined as \(L = 6.022 \times 10^{23}\text{ mol}^{-1}\).
Particle Number Relationship
\[ N = n \times L \]
Where \(N\) is the total number of particles, \(n\) is amount in moles, and \(L = 6.022 \times 10^{23}\text{ mol}^{-1}\).
2. Preparing a Standard Volumetric Solution
A solution of accurately known concentration prepared using a primary chemical standard.
Standard Laboratory Protocol (RP1 Alignment)
- Weigh by Difference: Weigh a weighing boat containing the primary standard (e.g. anhydrous \(\text{Na}_2\text{CO}_3\) or \(\text{H}_2\text{C}_2\text{O}_4\)) on a 2-decimal place analytical balance. Transfer the solid to a clean \(250\text{ cm}^3\) glass beaker. Reweigh the empty boat to account for any residual solid: \[ m(\text{transferred}) = m(\text{boat + solid}) - m(\text{empty boat}) \]
- Dissolution: Add approximately \(100\text{ cm}^3\) of deionised water to the beaker. Stir thoroughly with a glass rod until all crystals dissolve completely.
- Quantitative Transfer: Pour the solution through a clean filter funnel into a \(250.0\text{ cm}^3\) volumetric flask. Rinse the beaker, glass rod, and funnel with deionised water, collecting all washings in the volumetric flask.
- Make up to Mark: Add deionised water until the water level approaches the graduation line. Use a dropping pipette to add the final drops until the bottom of the curved meniscus rests exactly on the horizontal graduation line at eye level.
- Homogenisation: Stopper the flask tightly and invert 10 to 15 times to ensure thorough mixing and uniform concentration throughout.
3. Volumetric Titration Protocols
Titrations accurately determine the concentration of an unknown solution by reacting it with a standard solution of known concentration.
Apparatus Preparation Rules
- Pipette: Rinse with deionised water, then rinse with the solution to be pipetted. Deliver exactly \(25.0\text{ cm}^3\) into a conical flask.
- Burette: Rinse with deionised water, then rinse with the titrant solution. Ensure the space below the tap is completely filled with solution and free of air bubbles before recording initial volume.
- Conical Flask: Rinse with deionised water ONLY. A conical flask is used rather than a beaker because its sloping walls prevent splashing during swirling. Placed on a white tile to observe sharp indicator colour changes.
Concordant Titres
Perform a preliminary rough titration, followed by accurate titrations. Concordant titres are results that agree within \(0.10\text{ cm}^3\) of each other. Only concordant titres are averaged when calculating the mean titre volume.
4. Back Titrations
A back titration is employed when an analyte is insoluble in water (such as \(\text{CaCO}_3\)), reacts too slowly with the titrant, or has no direct visual endpoint.
Step 1: Calculate moles of NaOH used in the titration:
\[ n(\text{NaOH}) = c \times V = 0.100 \times (22.40 / 1000) = 0.002240\text{ mol} \]
Step 2: Find moles of unreacted HCl in the 25.0 cm^3 aliquot:
Reaction: \(\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}\) (1:1 ratio)
\[ n(\text{HCl in 25 cm}^3) = 0.002240\text{ mol} \]
Step 3: Scale up to find total unreacted HCl in the 250.0 cm^3 volumetric flask:
\[ n(\text{HCl unreacted total}) = 0.002240 \times \left(\frac{250.0}{25.0}\right) = 0.02240\text{ mol} \]
Step 4: Calculate initial moles of HCl added:
\[ n(\text{HCl initial}) = 1.000 \times (50.0 / 1000) = 0.05000\text{ mol} \]
Step 5: Calculate moles of HCl that reacted with CaCO3:
\[ n(\text{HCl reacted}) = 0.05000 - 0.02240 = 0.02760\text{ mol} \]
Step 6: Calculate moles and mass of pure CaCO3:
Reaction: \(\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2\)
\[ n(\text{CaCO}_3) = n(\text{HCl reacted}) / 2 = 0.02760 / 2 = 0.01380\text{ mol} \]
\[ m(\text{pure CaCO}_3) = 0.01380 \times 100.1 = 1.38138\text{ g} \]
Step 7: Calculate percentage purity of the sample:
\[ \% \text{ Purity} = \frac{\text{Mass of pure }\text{CaCO}_3}{\text{Mass of impure sample}} \times 100 = \frac{1.38138\text{ g}}{1.500\text{ g}} \times 100 = \mathbf{92.1\%} \]
Reported to 3 significant figures: 92.1%.
5. Net Ionic Equations with State Symbols
In aqueous reactions, soluble ionic substances dissociate into individual hydrated ions. Net ionic equations omit non-reacting spectator ions to focus on the actual chemical transformation:
Step-by-Step Writing Rule
- Write full balanced stoichiometric equation with state symbols.
- Split only strong electrolytes in aqueous solution (\((aq)\)) into ions. Leave solids \((s)\), liquids \((l)\), and gases \((g)\) intact.
- Cancel identical spectator ions appearing on both sides.
- Ensure charges and atoms balance.
Full: \(\text{Na}_2\text{CO}_3(aq) + 2\text{HCl}(aq) \rightarrow 2\text{NaCl}(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)
Ionic: \(2\text{Na}^+(aq) + \text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) + 2\text{Cl}^-(aq) \rightarrow 2\text{Na}^+(aq) + 2\text{Cl}^-(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)\)
Net Ionic: \[\mathbf{\text{CO}_3^{2-}(aq) + 2\text{H}^+(aq) \rightarrow \text{H}_2\text{O}(l) + \text{CO}_2(g)}\]
6. Practice Questions
Solution:
\[ \mathbf{\text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \rightarrow \text{BaSO}_4(s)} \]
(1 mark for correct species and balance; 1 mark for correct state symbols).