Aqueous SolutionsConcentration - Calculations.ppt

1、Aqueous Solutions Concentration / Calculations,Dr. Ron Rusay Spring 2002, Copyright 1995-2002 R.J. Rusay,Solutions,Homogeneous solutions are comprised of solute(s), the substance(s) dissolved, The lesser amount of the component(s) in the mixture, and solvent, the substance present in the largest amo

2、unt. Solutions with less solute dissolved than is physically possible are referred to as “unsaturated”. Those with a maximum amount of solute are “saturated”. Occasionally there are extraordinary solutions that are “supersaturated” with more solute than normal.,Concentration and Temperature,Relative

3、 Solution Concentrations: Saturated Unsaturated Supersaturated,DHMO, dihydromonoxide : “The Universal” Solvent,http:/www.dhmo.org,Water : “The Universal” Solvent,Generally, likes dissolve likes, i.e. polar-polar and nonpolar-nonpolar. Polar and nonpolar do not mix , eg. oil and water,The oil (nonpol

4、ar) and water (polar) mixture dont mix and are immiscible. If liquids form a homogeneous mixture, they are miscible.,Aqueous Reactions & Solutions,Many reactions are done in a homogeneous liquid or gas phase which generally improves reaction rates. The prime medium for many inorganic reactions is wa

5、ter which serves as a solvent (the substance present in the larger amount), but does not react itself. The substance(s) dissolved in the solvent is (are) the solute(s). Together they comprise a solution. The reactants would be the solutes. Reaction solutions typically have less solute dissolved than

6、 is possible and are “unsaturated”.,Salt dissolving in a glass of water,Solution Concentration,A solutions concentration is the measure of the amount of solute dissolved. Concentration is expressed in several ways. One way is mass percent. Mass % = Mass solute / Mass solute + Mass solvent x100 What

7、is the mass % of 65.0 g of glucose dissolved in 135 g of water?Mass % = 65.0 g / 65.0 + 135g x 100= 32.5 %,Solution Concentration,Concentration is expressed more importantly as molarity (M).Molarity (M) = Moles solute / Liter Solution An important relationship is M x Vsolution= mol This relationship

8、 can be used directly in mass calculations of chemical reactions. What is the molarity of a solution of 1.00 g KCl in 75.0 mL of solution? M KCl = 1.00g KCl / 75.0mL 1molKCl / 74.55 g KCl 1000mL / L= 0.18 molKCl / L,Solution Concentration,The following formula can be used in dilution calculations:M1

9、V1 = M2V2 A concentrated stock solution is much easier to prepare and then dilute rather than preparing a dilute solution directly. Concentrated sulfuric acid is 18.0M. What volume would be needed to prepare 250.mL of a 1.50M solution? V1 = M2V2 / M1 V1 = 1.50 M x 250. mL / 18.0 M V1 = 20.8 mL,Solut

10、ion Applications,A solution of barium chloride was prepared by dissolving 26.0287 g in water to make 500.00 mL of solution. What is the concentration of the barium chloride solution? MBaCl2 = ? MBaCl2 = = 26.0287g BaCl2 / 500.00mL 1mol BaCl2 / 208.23g BaCl2 1000mL / L,= 0.25000 mol / L,Solution Appl

11、ications,10.00 mL of this solution was diluted to make exactly 250.00 mL of solution which was then used to react with a solution of potassium sulfate. What is the concentration of the diluted solution. M2 = ? MBaCl2 = M1 M2 = M1V1 / V2 M2 = 0.25000 M x 10.00 mL / 250.00 mL M2 = 0.010000 M,Solution

12、Applications,20.00 mL of the barium chloride solution required 15.50 mL of the potassium sulfate solution to react completely. MK2SO4 = ? BaCl2(aq) + K2SO4(aq) ? + ?BaCl2(aq) + K2SO4 (aq) 2 KCl(aq) + BaSO4 (s) ?MK2SO4 = MBaCl2x VBaCl2 / VK2SO4 ? molK2SO4 / ? molBaCl2 0.010000 molBaCl2 x 0.02000 LBaC

13、l2 x 1 molK2SO4 LBaCl2 x 0.01550 LK2SO4 x 1 molBaCl2,?MK2SO4 =,?MK2SO4 = 0.01290 molK2SO4 / LK2SO4 = 0.01290 MK2SO4,Solution Applications,How many grams of potassium chloride are produced? BaCl2(aq) + K2SO4(aq) ? + ? BaCl2(aq) + K2SO4 (aq) 2 KCl(aq) + BaSO4 (s),?gKCl = 0.010000 molBaCl2 / LBaCl2 x 0

14、.02000 LBaCl2 X 2 molKCl / 1 molBaCl2 X 74.55 gKCl /molKCl,1,= 0.02982 gKCl,Solution Applications,If 20.00 mL of a 0.10 M solution of barium chloride was reacted with 15.00 mL of a 0.20 M solution of potassium sulfate, what would be the theoretical yield of barium sulfate?BaCl2(aq) + K2SO4 (aq) 2 KC

15、l(aq) + BaSO4 (s),MolBaCl2 = MBaCl2x VBaCl2 = 0.10 molBaCl2 / LBaCl2 x 0.02000 LBaCl2 1 molBaCl2= 2.0 x 10-3,MolK2SO4= MK2SO4x VK2SO4 = 0.20 molK2SO4 / LK2SO4 x 0.01500 LK2SO4 1 molK2SO4= 3.0 x 10-3,Which is the Limiting Reagent?,2.0 x 10-3 3.0 x 10-3,2.0 x 10-3 mol is limiting,Solution Applications

16、,If 20.00 mL of a 0.10 M solution of barium chloride was reacted with 15.00 mL of a 0.20 M solution of potassium sulfate, what would be the theoretical yield of barium sulfate?BaCl2(aq) + K2SO4 (aq) 2 KCl(aq) + BaSO4 (s),Must use the limiting reagent: 0.10 molBaCl2 x 0.02000 LBaCl2 x 1 molBaSO4 x 233.39 g BaSO4LBaCl2 1 molBaCl2 molBaSO4,=,= 0.47 g,