VOLUMETRIC ANALYSIS In volumetric analysis, the amount (in mol) of the substance being analysed is determined by measuring the volume of the test solution required to react completely with a volume of solution of known concentration. •Concentration is a measure of the relative amounts of solute and solvent. It is measured in molarity: mol L-1 or M. •Concentration = amount of solute, mol Volume of solution, L •To convert a concentration measured in mol L-1 to (g L-1), simply multiply the molarity by the molar mass. UNITS ARE THE KEY! •Eg. 2.00 mol L-1 NaCl Mol x g = g L mol L 2.00mol L-1 x (23.0 + 35.5) g mol-1 = 117 g L-1 STANDARD SOLUTIONS •Solutions with accurately known concentrations are called standard solutions. Preparing a standard solution is not a simple matter of dissolving a measured mass as many chemicals are impure as they decompose or react with chemicals in the atmosphere. •Substances that are so pure that the amount of substance, in mole, can be calculated accurately from their mass are called Primary Standards. They should: •Be readily obtainable in pure form •Have a known formula •Be easy to store without decomposing or reacting with the atmosphere •Have a high molar mass to minimise the effect of errors of weighing •Be inexpensive •E.g.’s Basic primary standard = Na2CO3 (anhydrous sodium carbonate), Na2B4O7.10H2O (sodium borate) • Acidic primary standard = H2C2O4.2H2O (hydrated oxalic acid). To make a standard solution: •Place weighed mass of solute in volumetric flask •Half fill with water and shake to dissolve •Add water to the calibration line. Make sure you eye is in line with this to avoid parallax errors. •Calculate the exact concentration of the standard. Solution stoichiometry •Chemical reactions involving solutions such as precipitation and neutralisation. •Formula to use: Solutions: n = C x V Sample Problem: •A solution of cloudy ammonia is analysed for its ammonium hydroxide concentration. A 20.00 mL sample is neutralised with 30.00 mL of 1.0 M hydrochloric acid. •Find the concentration of the ammonium hydroxide in the cloudy ammonia. Step 1: Write a balanced equation and identify known and unknown. NH4OH (aq) + HCl (aq) → NH4Cl(aq) + H2O (l) unknown known V = 20.00 mL V = 30.00 mL C = ??? C = 1.0 M Step 2: Calculate n (known). n(HCl) = C x V = 1.0 x 0.0300 = 0.030 mol Step 3: From equation find ratio of n(unknown) to n(known). n(NH4OH) = 1/1 x n(HCl) = 1/1 x 0.030 = 0.030 mol Step 4: Answer Question Find concentration of ammonium hydroxide. C (NH4OH) = n / V = 0.030 / 0.0200 = 0.15 M |
Volumetric flask used in making up a standard solution.
Diagram courtesy of E. Generalic, http://glossary.periodni.com/glossary.php?en=volumetric+flask LIMITING REAGENTS
•A limiting reactant is completely used up in the chemical reaction. •The other reactants are known as excess reactants and are not completely used up. Sample Problem: 20.0 mL of a 1.00M LiOH solution is added to 30.0mL of a 0.500M HNO3 solution. What is the limiting reagent and the mass of LiNO3 contained in the mixture? Step 1: Write a balanced equation and identify known and unknown. LiOH (aq) + HNO3(aq) -->LiNO3 (aq) + H2O (l) 1.00M 0.500M m=? 20.0mL 30.0mL Step 2: Calculate n (known). n(LiOH) = cxV = 1.00 x 0.0200 = 0.0200 mol. n(HNO3) = cxV = 0.500 x 0.0300 = 0.0150 mol. Compare mole ratio from equation: LiOH : HNO3 1 : 1 0.0200 : 0.0150 Comparing the moles of each- HNO3 will be completely used up in the reaction- so this means that HNO3 is the limiting reagent. (This is used to find mass of LiNO3) LiOH is in excess by 0.0200-0.0150 = 0.0050 moles. Step 3: From equation find ratio of n(unknown) to n(known). n(LiNO3) = 1/1 x n(HNO3) = 1/1 x 0.0150 = 0.0150 mol Step 4: Answer Question Find mass of lithium nitrate. m (LiNO3) = n x M = 0.0150 x 68.9 = 1.03 g |