Concentration of Solutions Molarity Two solutions can contain

Concentration of Solutions

Molarity • Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. • Molarity is one way to measure the concentration of a solution. moles of solute Molarity (M) = volume of solution in liters

Mixing a Solution Procedure for preparing 0. 250 L of 1. 00 M solution of Cu. SO 4 • Weight out 39. 9 g of Cu. SO 4 = 0. 250 mol • Add 250 m. L of water to volumetric flask • Molarity =. 250 mol/. 250 L = 1. 00 M

Expressing the concentration of an Electrolyte • When an ionic compound dissolves, the relative concentrations of the ions introduced into the solution depends on the chemical formula of the compound – Example: • 1. 0 M Na. Cl – 1. 0 M Na+ – 1. 0 M Cl- • 1 M Na 2 SO 4 – 2. 0 M Na+ – 1. 0 M SO 42 -

Using Molarities in Stoichiometric Calculations

Practice Question • How many moles of HNO 3 are in 2. 0 L of 0. 200 M HNO 3 solution? Moles = 2. 0 L soln x 0. 200 mol HNO 3 / 1 L solution =0. 40 mol HNO 3

Dilution

Dilution • Solutions of lower concentration can be obtained by adding water • Moles solute in conc soln = moles solute in dilution sln • Therefore, Mconc x Vconc = Mdil x Vdil

Practice problem • How many liters of 3. 0 M H 2 SO 4 are needed to make. 450 L of 0. 10 M H 2 SO 4? Mconc x Vconc = Mdil x Vdil Moles H 2 SO 4 in dilute solution = 0. 450 L x 0. 10 mol/L = 0. 045 mol = moles H 2 SO 4 in conc solution L conc soln = 0. 045 mol x 1 L/ 3. 0 mol =. 015 L

Titration • The analytical technique in which one can calculate the concentration of a solute in a solution. • One reagant has known concentration

Titration • Indicators are used to determine the equivalence point of the reaction – Point where the neutralization reaction between 2 reactants are complete – Reactants in stoichiometric equivalents are brought together

Practice Problem If 45. 7 m. L of 0. 500 M H 2 SO 4 is required to neutralize a 20. 0 m. L sample of Na. OH solution. What is the concentration of the Na. OH solution? Moles H 2 SO 4 =. 0457 L x 0. 500 mol H 2 SO 4 /L = 2. 28 x 10 -2 mol H 2 SO 4 Balanced Equation: H 2 SO 4 + 2 Na. OH 2 H 2 O + Na 2 SO 4 Moles Na. OH = 2. 28 x 10 -2 mol H 2 SO 4 x 2 mol Na. OH/1 mol H 2 SO 4 = 4. 56 x 10 -2 mol Na. OH Molarity of Na. OH = 4. 56 x 10 -2 mol Na. OH/. 020 L soln = 2. 28 M Na. OH