I got a 1st in my first year studying chemistry at the University of Birmingham using these revision notes that I have uploaded. They include detail on gravimetric analysis, the solubility product, common ion effect, volumetric analysis, the operation of indicators, acid-base titrations, calculatin...
Acids, Bases, Buffers and Solubility
25 September 2017 15:29
Gravimetric Analysis
GRAVIMETRIC ANALYSIS, PRINCIPLES AND PRACTICE • The mass of the product which has been formed is used to calculate the quantity of the original
• Convert between Ka and pKa, [H+] and pH analyte (species being analysed/substance whose chemical constituents are being identified
• Interconvert solubilities and solubility products and measured)
• An analyte is reacted to form a solid of precisely known composition
• In combustion analysis, a sample is burned in excess oxygen and • The mass of the solid produced allows the concentration of analyte to be determined.
products are measured. Typically used to measure C, H, N, S and e.g. BaSO4 precipitation: Ba2+(aq) + SO42-(aq) => BaSO4(s)
halogens in organic matter. Solid is filtered and dried, what is the percentage of Ba in BaSO4?
• To measure other elements, organic matter is burned in a closed RMM Ba = 137.327, RMM BaSO4 = 233.3896. %Ba = (137.327/233.3896)x100 = 58.8402 %
system. Products then dissolved in acid/base and measured by • The precipitate maybe converted into a compound with a more precise stoichiometry known as
inductively coupled plasma with atomic emission or mass spec a weighing form
Solubility Product
• The solubility product is the equilibrium constant for the reaction in Must know the exact % of the analyte in the weighing form.
which a solid salt dissolves to give its constituent ions in solution.
• A "supersaturated" solution contains more solute than should be For Co(OH)2: Co(OH)2(s) ⇌ Co2+(aq) + 2OH-(aq), KSP=[Co2+][OH-]2
present at equilibrium If [Co2+][OH-]2 >KSP then Co(OH)2 precipitates.
Consider the determination of Cl- ions by precipitation with excess Ag+ to For AgCl, KSP=1.8x10-10 at 25oC Therefore solubility (molL-1) is 1.34x10-5 molL-1
explore background principles: What is the solubility of AgCl in water at 25oC? RMM AgCl = 143.321
Solubility relates to the equilibrium AgCl(s) ⇌ Ag+(aq) + Cl-(aq) Solubility (gL-1) = 1.34 x 10-5 x 143.321 = 1.92 x
(Note AgCl is solid so [AgCl]=1, solid emitted from equilibrium as it is in its KSP = [Ag+][Cl-] and [Ag+] = [Cl-] (1:1 stoichiometry) 10-3gL-1
standard state) KSP = 1.8 x 10-10 = [Ag+]2 If we add excess precipitant (Ag+), then [Ag+]
KSP is known as the solubility product [Ag+] = √KSP = 1.34 x 10-5 molL-1 will be higher than the simple saturated value
Example What is solubility if [Ag+] = 0.1M?
A 10mL solution containing Cl- ions was treated with excess AgNO3 and the • The product [Ag+][Cl-] is constant at equilibrium in the presence of excess solid AgCl.
dried AgCl weighed 0.4368g. What was the molarity of Cl- in the solution? • The Ag+ in solution from the dissolved AgCl is negligible compared with that from the excess
RMM AgCl = 143.321 AgNO3 added - solubility given by [Cl-]:
0.4368g AgCl is 3.05x10-3mol
Hence 10mL contains 3.05 x 10-3 mol so 1L contains
Solution is 0.305M in Cl- AgCl solubility is 1.8 x 10-9 x 143.321 = 2.58 x 10-7gL-1 (solubility gone from 1.92x10-3 to 2.58 x
10-7gL-1, solubility reduced by ~10-4)
Problem: Gravimetric Analysis This is known as the common ion effect. Le Chatelier's principle "a salt will be less soluble if one of
0.3516g of a phosphate detergent was ignited at red heat to destroy the its constituent ions is already present in the solution".
organic matter. The residue was dissolved in hot HCl and the phosphate was AgCl(s) ⇌ Ag+(aq) + Cl-(aq). Excess Ag+ has moved the equilibrium to the left, adding excess
converted to H3PO4. The phosphate was precipitated as MgNH4PO4.6H2O by Ag+ or Cl- reduces the solubility of AgCl.
addition of excess Mg2+ followed by aqueous NH3. The precipitate was filtered
and washed, and then converted to a stable weighing form, Mg2P2O7 by Common ion effect problems
igniting at 1000oC. The residue weighed 0.2161g. Calculate the percentage P in 1. Calculate the solubility of PbCl2 at 25oC in water and in a solution which is 0.0646M in Pb2+
the original sample. (KSP = 9.8x10-9)
• In 0.0646M Pb2+
RMM Mg2P2O7 = 222.5534, RAM P = 30.9738 • In 25oC water: PbCl2(s) ⇌ Pb2+(aq) + 2Cl-(aq)
Amount of P in the weighed product is: 0.2161 x (2xRAM(P))/RMM(Mg2P2O7) =
0.0602g
%P in sample = (0.0602/0.3516)x100 = 17.1217 %
Applicability
• Often high precision (high agreement between sets of results)
• High accuracy (results close to true answer)
• Poor sensitivity (measure of lowest concentration that can be measured to 2. Calculate the solubility of AgCl at 25oC in water and in a solution
acceptable precision) which is 7.97x10-3M in Ag+ (KSP = 1.77x10-10).
○ This is because it relies on solubility • In 25oC water: AgCl ⇌ Ag+ + Cl- • In 7.97x10-3M Ag+
Good for major components
• Selectivity (detects and measures just one species) is sometimes poor e.g. Ba,
Sr, Ca all precipitated by SO42-
Improvements to accuracy and precision
• Good practice (appropriate weighing form, heat to constant weight (use desiccator,
avoid touching))
• Enhance precipitate properties (increase size (filterability - ability to catch everything by Examples of gravimetric analysis
filtration). Ways to increase size are via digestion (redissolve and reprecipitate) and via Inorganic precipitants e.g. Cl- (Ag+), SO42- (Ba2+), OH- (Fe3+)
PFHS (precipitation from homogeneous solution) Organic precipitants are generally chelating molecules:
○ PFHS generates the precipitant slowly in situ e.g. dimethylglyoxmine (DMG)
PFHS example: DMG is highly specific for Ni2+, Pd2+, Pt2+ i.e. metals which form square planar
Urea to form OH- e.g. in precipitation of Al, Fe hydroxides. CO(NH2)2 + 3H2O => CO2 + 2NH4+ + complexes
2OH-, reflux 1-2 hours with Fe3+ (Fe3+ and OH- react to form precipitate)
VOLUMETRIC ANALYSIS - PRINCIPLES
Compared to Fe3+ + 3NH4OH => Fe(OH)3 • Describe the operation of indicators
+ 3NH4+, much more rapid precipitation • A titration is a type of volumetric analysis during which increments of a reagent
Produces larger particles easier to filter. More difficult to capture through normal solution (titrant - delivered from a burette) are added to an analyte until their
Crystalline precipitate, not fine powdery filtration process. reaction is complete
or gelatinous. • The quantity of analyte originally present is calculated using the quantity of
titrant required to complete the reaction
Difference between equivalence point and endpoint should be small - causes error. Can be • Equivalence point - quantity of titrant is precisely correct for a given
allowed for by blank titration - same procedure but without analyte. The blank titration value stoichiometric reaction with analyte. The ideal theoretical result in a titration.
subtracted from value with analyte. Reactions are monitored by physical quantity (e.g. colour) that changes abruptly
• Endpoint cannot exactly equal equivalence point because extra e.g. MnO4- - some small at equivalence point
amount beyond that needed to react, is required to exhibit purple colour. • Indicator - compound added to cause such a colour change
Titrants must have well-defined volume and concentration • Endpoint - point of the abrupt (colour) change signalling complete reaction with
• Volume: class A - calibrated individually; used in "quality assured labs", class B - analyte.
satisfactory for routine use
• Concentration: primary standard - exceptionally stable solids e.g. Na2CO3 for a base ACID-BASE TITRATIONS
(cannot normally use), secondary standard - calibrated first against a primary standard • Calculate the pH at all points in an acid-base titration
e.g. HCl may be calibrated against Na2CO3 as its concentration can vary. • Understand and be able to use the Henderson-Hasselbalch equation
Inorganic Chemistry I Page 1
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