Date post: | 22-Dec-2015 |
Category: |
Documents |
Upload: | barbra-taylor |
View: | 215 times |
Download: | 2 times |
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 litersMolarity (M) =
Mixing a Solution
Procedure for preparing 0.250 L of 1.00 M solution of CuSO4
• Weight out 39.9 g of CuSO4
= 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 Na2SO4
– 2.0 M Na+
– 1.0 M SO42-
Using Molarities inStoichiometric Calculations
Practice Question
• How many moles of HNO3 are in 2.0 L of 0.200 M HNO3 solution?
Moles = 2.0 L soln x 0.200 mol HNO3 / 1 L solution
=0.40 mol HNO3
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 H2SO4 are needed to make .450 L of 0.10 M H2SO4?
Mconc x Vconc = Mdil x Vdil
Moles H2SO4 in dilute solution = 0.450 L x 0.10 mol/L
= 0.045 mol = moles H2SO4 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 ProblemIf 45.7 mL of 0.500 M H2SO4 is required to neutralize a
20.0 mL sample of NaOH solution. What is the concentration of the NaOH solution?
Moles H2SO4 = .0457 L x 0.500 mol H2SO4 /L
= 2.28 x 10-2 mol H2SO4
Balanced Equation: H2SO4 + 2 NaOH 2 H2O + Na2SO4
Moles NaOH = 2.28 x 10-2 mol H2SO4 x 2 mol NaOH/1 mol H2SO4
= 4.56 x 10-2 mol NaOH
Molarity of NaOH = 4.56 x 10-2 mol NaOH/.020 L soln= 2.28 M NaOH