Solutions - concentrations, titration calculations and back titration
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| A solution is | a homogenous mixture of two or more substances
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| The solvent is | the liquid, present in excess, in which dispersion occurs
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| The solute is | the substance dissolved in the solvent, can be a solid, liquid or gas.
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| Solutions are different from suspensions because... | solutions are transparent, do not settle out and cannot be separated by filtration
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| A suspension is | when fine particles of a solid are in a liquid
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| Solubility = | the quantity of that substance that will dissolve to form a solution
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| Solubility varies with | temperature - with solids increases w/ temp, with gases decreases w/temp
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| Concentrated solution = | when a solution contains a large amount of solute
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| Dilute solution = | when a solution contains a small amount of solute
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| Saturated solution = | solution in which no more solute will dissolve at that temperature
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| Precipitate = | excess solid that eventually separates from a saturated solution
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| Concentration = | the amount of a substance (moles) contained within a given volume of solution.
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| 1dm3 = | 1 litre
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| Calculating concentration = | moles of solute / volume in dm3
i.e. C = n/V (n = c x v)
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| Ionic substances split up into their component ions when dissolved in water therefore.... | the concentrations of the individual ions will depend on how many of these ions are produced when the substance dissolves.
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| Standard solution = | solution of accurately known concentration used in titration
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| Primary standard = | a solution prepared by dissolving a precisely KNOWN MASS of a PURE solute to make an accurately KNOWN VOLUME of solution using a volumetric flask. (use to check concentration of other solution used in titration)
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| A primary standard must: (5) | - be stable in the form of the solid and solution
- be available in very pure form
- relatively high molar mass
- readily soluble in water
- react COMPLETELY in KNOWN manner
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| Examples of primary standards: | potassium hydrogenphthalate and sodium carbonate
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| How titration works | - accurately known volume of one solution is pipetted into conical flask
-indicator usually added
- other solution with is run into conical flask using a burette until indicator just changes colour
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| Calculating titre | subtract initial burette reading from final one.
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| Calculating amount of solute (moles) | by using the volume of solution of known concentration
(C x V)
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| Calculating amount of unknown solute (moles) | by using balanced chemical equation
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| Calculating concentration of unknown solute | by using the number of moles previously calculated from equation and the volume of the second solution used (titre).
( C = n/V )
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| Back titration is used when | the reaction occurs too slowly for titration. Usually when insoluble solid reagents are used.
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| Back titration | 1. a reaction is carried out b/w insoluble base and excess of known volume of standard solution of acid
2. then this is titrated with an alkali of known concentration to then determine number of moles of the reagent in excess that was left unreacted.
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| Back titration 2 | 3. then by knowing the initial moles of reagent and the moles of remaining excess reagent you can find out the moles that has reacted with the sample.
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| Total (known) amount of acid | is the sum of the amount that reacted with the known alkali and the amount that reacted with the sample (unknown).
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| When calculating the back titration | always start with the reaction between the acid and the base first! to calculate the actual moles of the acid that reacted.
then calculate the mol of acid that reacted w/the insoluble base by subtracting mol that reacted w/ acid from mol of initial acid
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| When calculating the back titration 2 | secondly use the equation of acid reacting with insoluble base to find out moles of insoluble base.
then find out mass!
woo
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