Concentration Molarity Normality Equivalence and Titrimetry Objective To

Concentration, Molarity, Normality, Equivalence and Titrimetry Objective To understand the common units of Quantity that are used in water analysis. To understand the reason why a standard and consistent approach is necessary in the way that measured values are expressed. To be able to calculate and express results in the relevant units.

Forms of measurement n Fundamental importance – mass concentration – molar concentration – normality and equivalent l used in standard solutions

Mass concentration n Weight/volume – mg/l = mass of substance volume of solution – n used in water Weight/weight – ppm or ppb = mass of substance mass of solution or matrix – n soil, or sludge or compost ppm = mg/l (only if density of liquid is 1) n n ie 1 litre = 1 Kg ppb = g/l (only if density of liquid is 1)

Molar concentrations – Molecular weight = sum of atomic weight –NH 3 – 1 – mole = 14 + (3 x 1) = 17 g/mol = molecular weight in g = gram atom Molar concentration M = moles per litre

Molar Concentration v Example w A solution of formaldehyde is required to preserve some biological specimens in the laboratory. Method 1 specifies the use of 2 M CH 2 O. Method 2 specifies 6 % (w/v) CH 2 O. w Which will require the greater quantity of CH 2 O? v Method 1. MW of CH 2 O = 30 g/mole 2 M = 2 moles per litre 2 x 30 g/l = 60 g required for each litre prepared. v Method 2. 6% (w/v) CH 2 O is 6 g in every 100 ml solution 6 g/100 ml = 60 g/1000 ml = 60 g required for each litre prepared. ANSWER - they both require the same amount.

Titrimetry v What is Titrimetry w Buret and Flask w Indicators – Phenolphthalein red (base) to Colourless (acid) w Titration to an End-Point – amount of Titrant = amount of analyte – amount must be defined in a convenient form – reactivity x volume = amount

Equivalent and Normal Solutions By definition ‘ A STANDARD SOLUTION is one whose strength is known. 1. amount of substance (moles) per unit volume - moles/l or molar (M) 2. amount of reactive species (Equivalents) per unit volume - equivalents/l or normal ( N ) Think of N as meaning ‘Equivalent in Reactive Strength’ HCl H+ + Cl 1 M 1 mole H+ per litre H 2 SO 4 2 H+ + SO 42 - 1 M 2 mole H+ per litre equal molarity does not give equal reactivity However, one equivalent of each substance in a unit volume will give equivalence in reactivity. litre HCl H+ + Cl. H 2 SO 4 2 H+ + SO 42 - 1 eq/l 1 mole H+ per litre how do we find the weight of a substance that gives one equivalent?

Equivalent Weight v Equivalent Weight ( EW ) of an acid is defined as: ‘that weight of the substance which contains 1 gram atom of available hydrogen ’ (1 mole of hydrogen (H) weighs 1 gram atom = atomic weight) v In General terms the EW of a substance is given by: molecular weight divided by z MWt/z where z is an integer 1, 2, 3 etc. and its value is determined by the context of the Chemical reaction (acids = number of H+ available) For Ionic Reactions and Precipitation Reactions w z is given by the charge on the ion. For Oxidation Reduction (Redox) Reactions w z is given by the number of electrons involved in the half reaction

Normal Solutions A 1 Normal Solution is a solution containing 1 Equivalent Weight of a substance per litre of volume. v For Acids Example: 1 N solution of HCl = MWt/z per litre = 36. 5/1 per litre = 36. 5 g HCl per litre 1 N solution of H 2 SO 4 = MWt/z per litre = 98/2 per litre = 49 g H 2 SO 4 per litre v For Bases (Alkali) z equals the number of moles of H+ ie. (HCl) that would react with 1 mole of the base. Example: Na. OH + H+ Na+ + H 2 O Ca(OH)2 + 2 H+ Ca 2+ + 2 H 2 O z =1 z =2 Equivalent weight of Na. OH is MW/z = 40/1 = 40 g/equiv Equivalent weight of Ca(OH)2 is MW/z = 74/2 = 37 g/equiv

Normal Solutions v Ionic Reactions (eg Precipitation Reactions) the value of z is based on the ion charge. Ca 2+ + CO 32 - Ca. CO 3(s) z = 2 Equivalent weight of Ca. CO 3 is MWt/z = 100/2 EWt = 50 g per equivalent Al 2(SO 4)3 2 Al 3+ + 3 SO 42 - z = 6 v Redox Reactions the Equivalent Weight is based on the change in the value of the Oxidation Number. EWt = MWt/(number of electrons taking part in the half reaction) O 2 + 4 H+ + 4 e- 2 H 2 O EWt = 32/4 = 8 g per equiv Example The COD Analysis method uses Potassium Dichromate (K 2 Cr 2 O 7 ) to oxidise the chemical constituents in the sample. Cr 2 O 72 - + 14 H+ + 6 e- 2 Cr 3+ + 7 H 2 O (half reaction) z = 6 EWt = 294/6 = 49 g/equiv

Normal Solutions v General Case concentrations are N Cs has units g. l-1 (Ew is the equivalent weight) Cs has units mg. l-1

Concentration in Terms of Substance v Care should be taken to compare like with like. w mg/l (NO 3 -N) w mg/l NO 3 w mg/l Ca. CO 3 w meq/l Ca. CO 3 v Dionex and Wet Chemical Analysis w standardize units for comparison of data with Drinking Water Standards v Urban Wastewater Treatment Directive standards use Total Nitrogen and Total Phosphorus

Concentration in Terms of Substance v Example Consider the analysis of NH 4 and NO 3 in a water sample 360 mg NH 4+ per litre 1240 mg NO 3 - per litre for comparison of the data we must standardise the units NH 4+-N/l = 360 mg NH 4+ /l x 14 g. N/18 g NH 4+ = 280 mg NH 4+-N/l NO 3 --N/l = 1240 mg NO 3 - /l x 14 g. N/62 g NO 3= 280 mg NO 3 --N/l units are now the same e. g. mg nitrogen (mg N) so they can be summed: 280 mg NH 4+-N/l + 280 mg NO 3 --N/l = 560 mg. N/l

Stoichiometry The relative molar ratios of reactants and products in a balanced equation. w Used to calculate how much of a reactant is required. – Chemical Manufacturing; Water Treatment Chemicals. v Example Hydrogen reacts with Carbon Monoxide to give Methane and Water. 1. Chemical symbols: H 2 + CO CH 4 + H 2 O 2. Balanced Equation 3 H 2 + CO CH 4 + H 2 O (Law of Conservation of Mass) Stoiciometry 3 hydrogen molecules to every carbon monoxide molecule. Mole ratio 1 mole hydrogen to each 1/3 mole of carbon monoxide. Moles of desired substance = (moles of known substance) x (mole ratio desired to known)

Moles - Example v A wastewater contains 10 mg/l ammonia as NH 3. v How much oxygen is required to oxidise the ammonia in 1 litre of wastewater to nitrate (NO 3 -). v Step 1. w Write balance Equation and the find mole ratio. v Step 2. w calculate moles of ammonia v Step 3. w calculate moles of oxygen v Step 4. w calculate mg of oxygen

Moles - Example (Solution) 1. NH 3 + 2 O 2 HNO 3 + H 2 O 1 mole + 2 moles 2. 10 mg NH 3 1 mole + 1 mole = 10 x 10 -3 g NH 3 moles of ammonia = 10 x 10 -3 g NH 3 / 17 g. mole-1 (molecular weight of ammonia is 17 g. mole -1) = 5. 9 x 10 -4 moles NH 3 3. O 2 required is: 5. 9 x 10 -4 moles x 2 moles/ 1 mole = 11. 8 x 10 -4 moles O 2 (2 moles oxygen for each mole ammonia, from reaction stoichiometry) 4. weight of O 2 is: 11. 8 x 10 -4 moles O 2 x 32 g. mole-1 = 0. 038 g Oxygen (molecular weight of oxygen is 32 g. mole -1) = 0. 038 g Oxygen(x 1000 mg. g ) = 38 mg Oxygen required -1