Practical Skills review. Titrations Acid + base – indicator will be used. Strong acid + weak base...
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Transcript of Practical Skills review. Titrations Acid + base – indicator will be used. Strong acid + weak base...
Practical Skills review
TitrationsAcid + base – indicator will be used.
• Strong acid + weak base– methyl orange– endpoint is orange
• Weak acid+ strong base – Phenolphthalein– endpoint is pale pink
Redox reactions – can be self indicating• With potassium manganate (VII):
– 5Fe2+ + 8H+ + MnO4- 5Fe3+ + Mn2+ + 4H2O
– Ethanedioic acid + MnO4‑ CO2 + Mn2+ (may need heat)
– MnO4- in burette
– Endpoint given by pink colour remaining
• With sodium thiosulphate, Na2S2O3(aq)– I2 + 2S2O3
2- 2I- + S4O62-
– thiosulphate in burette– starch added when iodine solution is pale yellow– goes blue black– endpoint when this colour goes– I2 might have been formed from I- in another reaction e.g.
analysis of bleach:2I- + OCl- + 2H+ I2 + Cl- + H2O
In all cases the balanced equations will be given, so mole ratios are available
Performing the titration• Filling up the burette
– no need to go to zero– use a funnel, remove the funnel– run some out before you use it to fill the nozzle.
• The pipette– used to transfer fixed volumes of a solution– fill to above the line– take off the filler– control volume with your thumb– transfer into flask by removing your thumb– don’t force the solution out!!!!
• The indicator (if used)– max 5 drops– permanent (20s) colour change
• Need to get colour change with one drop of solution– Do a quick rough– Perform 2 more to get concordant results (within 0.1 cm3)
Recording values• table will be needed• showing initial and final values a long with the titre• 2 d.p., with 0 or 5 in second d.p.
– on a line = 0 e.g. 12.40 cm3
– between lines = 5 e.g. 22.65 cm3
Keep to a max of 3 titrations – 1 rough, 2 accurate; any more and time will be an issue.
rough Expt 1 Expt 2
Final/cm3
Initial/cm3
Titre/cm3
Titre values• the difference between 2 values e.g.
26.40 – 1.55 = 24.85 cm3
• Need to have 2 within 0.1 cm3 of each other e.g.24.85 and 24.75 cm3 for a mark.
• Take an average of your best titres– usually 2 or 3 values– rough can be used if it appears to match well with
another value– show which titre values you use.
• Error in a single reading– half the smallest unit– which is ½ x 0.1 = 0.05 cm3
– NB if 2 readings are used the error actually becomes 2 x 0.05 = 0.1 cm3
Possible calculations:
• Find concentration of an acid or base;
• Find molar mass of a reactant or part of a reactant e.g. the metal present in a carbonate or the n(H2O) present in a hydrated solid.
Layout your calculations in neat, clear and logical steps
NB an endpoint can also be determined by measuring the temperature rise of the reaction– small portions of acid maybe added to the base– the temperature rise is recorded; – when the reaction is complete the temperature
will fall;– 2 distinct lines/curves are plotted;– Where they intersect gives the endpoint
Weighing of a solidNeed to know a range of masses:• mass of weighing vessel e.g. plastic cup• mass of vessel + solid• mass of vessel + any residue (not always needed)• must be tabulated• be aware that it might need to be done twice in a single
table:
Masses/g Expt 1 Expt 2
Vessel
Vessel + solid
Vessel + residue
Solid added
This might be used for an enthalpy calculation e.g., or for a dehydration of a hydrated salt e.g. Na2CO3.xH2O
Enthalpy calculations• Solid reacting with a solution e.g. carbonate + acid.• Mass of solid determined as described.• Solution will be measured using a burette, pipette or more
likely a measuring cylinder (systematic error is on the cylinder; random error in reading is half the smallest division).
• Initial temperature will be measured by leaving the thermometer in the solution for a few minutes whilst the solid is being weighed out
• table will be needed to represent this data – again be aware that 2 trials may be required:
Temps/oC Expt 1 Expt 2
Final
Initial
∆T
Error is again half the smallest division usually, so 0.5oC.
CalculationH = m x c x ∆T• volume of solution (cm3)will normally give
the mass, m (g);• c, specific heat capacity of water is given
in the question (4.3 Jg-1oC-1);• ∆T is the average change in T found;• Gives a value in Joules• Convert to a kJmol-1 value
– divide by the number of moles of substance not in excess (usually the solid)
– divide by 1000.
Dehydrations
Solid, e.g. CuSO4.xH2O:– Weigh a boiling tube;– Add solid;– Reweigh boiling tube +
solid– Table required
Solid is then heated to drive off the water of crystallisation– boiling tube is reweighed– Process is repeated until
weight is constant – change is <0.05 g
– Table is required:
Mass/g
Boiling tube
Boiling tube + hydrated solid
Boiling tube + solid after heating
Anhydrous solid
Large box for several readings
Calculation:
• Mass of water = hydrated solid – anhydrous solid
• Convert masses of water and anhydrous solid to moles
• Get simplest ratio of solid : water to find x (CuSO4:H2O)
• If x is already known it can be used to find the molar mass of other parts of the formula e.g. XCO3.5H2O
Precipitations
• Methods, hints and ideas provided very clearly in the lab manual.
• New bit – looking at halides– need to separate off the silver halide ppt before adding
NH3 to confirm Cl, Br or I.– Reason? – if Cu2+ or Fe3+ is present a ppt of the
hydroxide would form as well as the possibility of the silver halide dissolving or not.
• Care also needed with coloured solutions– Colour of ppt vs colour of the solution– need to let ppt settle or separate it out by decanting
and washing or filtration.
The writing
• Clear descriptions of tests to be performed are required e.g.– Add a few drops of NaOH followed by an excess– Add 1 cm depth of silver nitrate followed by an excess
of NH3.
• Clear statements of observations are needed– ppt or not– colour of ppt – including white– does it dissolve in excess reagent or not– is a gas given off? And if so how is it identified?