Qualitative inorganic analysis Flashcards
Hydroxides trend down group 2
Solubility increases down the grouo
Carbonates trend down group 2
No trend
Sulphates trend down group 2
Solubility decreases down group 2
Test to identify ammonium
NaOH, gently heat, hold damp red litmus over
Production of ammonia gas, litmus turns blue
Test to identify carbonate
HCl/H2SO4, test gas with calcium hydroxide solution (limewater)
Effervescence, CO2 gas produced, limewater turns cloudy
Test to identify sulfate
HCl then barium chloride
Production of white precipitate
Test to identify chloride (aq)
Nitric acid, then silver nitrate then add dilute ammonia solution
Production of white precipitate, precipitate dissolves
Test to identify chloride (solid)
Add concentrated H2SO4
Production of steamy fumes of HCl
Test to identify bromide (aq)
Nitric acid, then silver nitrate then add concentrated ammonia solution
Production of cream precipitate and precipitate dissolves
Test to identify bromide (solid)
Add concentrated H2SO4
Production of steamy fumes HBr, and red/brown Br2, and pungent smelling SO2
Test to identify iodide (aq)
Nitric acid then silver nitrate then concentrated ammonia solution
Production of pale yellow precipitate and no visible change
Test to identify iodide (solid)
Add concentrated H2SO4
Production of steamy fumes of HI, H2S (smells of rotten eggs), black sold iodine and sometimes yellow sold sulfur
Test to identify thiosulphate ion
Add sulphuric/hydrochloric acid
Yellow precipitate forms
Test to identify lead (II)
Add potassium iodide solution
Bright yellow/canary yellow ppt formed
Addition of OH- and then excess OH- to [Cu(H2O)6]2+
Blue precipitate
No visible change
Addition of OH- and then excess OH- to [Fe(H2O)6]2+
Green precipitate
No is visible change
Addition of OH- and then excess OH- to [Cr(H2O)6]3+
Grey precipitate
Precipitate dissolves forming a green solution
How do we go from gdm^-3 to moldm^-3?
To moldm^-3 = multiply by Mr
The other way is divide
Factors to consider when using a method, apart from cost
Rate of reaction
Atom economy
Temperature
Catalysts
Formation of side products
Why can’t the molar % of each compound be used to find the percentage by mass of each component if the relative molecular mass of each component differs?
If relative molecular masses are different, the number of moles present in the chromatogram is not directly related to the mass of each component
