 # Concentration of Solutions. Molarity Two solutions can contain the same compounds but be quite different because the proportions of those compounds are.

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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 volume of solution in liters Molarity (M) =

Mixing a Solution Procedure for preparing 0.250 L of 1.00 M solution of CuSO 4 Weight out 39.9 g of CuSO 4 = 0.250 mol Add 250 mL 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 NaCl –1.0 M Na + –1.0 M Cl - 1 M Na 2 SO 4 –2.0 M Na + –1.0 M SO 4 2-

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

Solutions of lower concentration can be obtained by adding water Moles solute in conc soln = moles solute in dilution sln Therefore, M conc x V conc = M dil x V dil

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 ? M conc x V conc = M dil x V dil 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 mL of 0.500 M H 2 SO 4 is required to neutralize a 20.0 mL sample of NaOH solution. What is the concentration of the NaOH 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 NaOH  2 H 2 O + Na 2 SO 4 Moles NaOH = 2.28 x 10 -2 mol H 2 SO 4 x 2 mol NaOH/1 mol H 2 SO 4 = 4.56 x 10 -2 mol NaOH Molarity of NaOH = 4.56 x 10 -2 mol NaOH/.020 L soln = 2.28 M NaOH

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