Basic Titrations

Working out unknown concentrations becomes straightforward once you've got the formula down. You'll use C = n/V where C is concentration (mol dm⁻³), n is moles, and V is volume (dm³).

The key is recognising that in a 1:1 reaction ratio, both solutions contain the same number of moles. Convert your volumes from cm³ to dm³ by dividing by 1000, then calculate moles of the known solution using n = CV.

Since the moles are equal, you can find the unknown concentration by rearranging C = n/V. In this example, 0.01 moles in 0.025 dm³ gives 0.4 mol dm⁻³.

Quick Tip: Always check your reaction equation to confirm the mole ratio - it's not always 1:1!

Double Titrations

Double titrations tackle mixtures containing two different compounds that react at different stages. You'll use two indicators to spot each endpoint clearly.

In the first stage with phenolphthalein, HCl neutralises all the NaOH plus half the Na₂CO₃ (22 cm³ used). The second stage with methyl orange completes the Na₂CO₃ reaction (8.25 cm³ more acid).

The clever bit is working backwards: the second titration tells you the total Na₂CO₃ present, whilst the first titration reveals the combined effect. Subtract to find the pure NaOH amount.

Calculate each concentration using C = (n × 1000)/V, giving you 0.055 mol dm⁻³ for NaOH and 0.033 mol dm⁻³ for Na₂CO₃.

Remember: Phenolphthalein changes first, then methyl orange - this sequence is crucial for double titrations!

Concentration Calculations

The final step in double titrations involves converting your mole calculations into proper concentration values. You'll need the formula C = (n × 1000)/V where n is moles and V is volume in cm³.

For the NaOH calculation, take your moles (1.375 × 10⁻³) and multiply by 1000, then divide by the original sample volume (25 cm³). This gives you 0.055 mol dm⁻³.

The Na₂CO₃ follows the same pattern: 8.25 × 10⁻⁴ moles converted gives 0.033 mol dm⁻³. These final answers tell you exactly what's in your original mixture.

Top Tip: Double-check your arithmetic here - small calculation errors can throw off your entire analysis!

Back Titrations

Back titrations are your go-to method when direct titration isn't practical, especially with insoluble samples or slow reactions. You add excess acid, then titrate backwards to find what wasn't used.

Start by calculating the excess HCl that didn't react with your sample. The NaOH titration (10.9 cm³) tells you exactly how much acid was left over - that's 2.485 × 10⁻³ moles.

Subtract this excess from your initial HCl amount (0.02 moles) to find what actually reacted with the Ba(OH)₂. Remember the 2:1 ratio in the equation - each Ba(OH)₂ needs two HCl molecules.

Finally, convert moles to mass using molar mass, then calculate percentage purity: $actual mass/sample mass$ × 100. Here, you get 90.8% purity.

Key Point: Back titrations are perfect when your sample doesn't dissolve well or reacts too slowly for direct methods!