Chapter 15 Solutions.

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Presentation transcript:

Chapter 15 Solutions

15.1 Solubility Solution: homogeneous mixture or mixture in which components are uniformly intermingled Solute: substance that is being dissolved in solvent Solvent: substance that dissolves solvent and present in a large amount Aqueous solutions: solutions with water as the solvent

Table 15.1 – Various Types of Solutions

15.2 Solution Composition: An Introduction Saturated: A solution in which the maximum amount of solvent has been dissolved. Any more solute added will sit as crystals on the bottom of the container Unsaturated: A solution in which more of solute can be dissolved Concentrated: a relative large amount of solute is being dissolved in solvent Diluted: a relative small amount of solute is being dissolved in solvent

15.3 Solution Composition: Mass Percent mass of solute Mass percent (m/m%) = ------------------------ x 100 mass of solution

Examples A solution is prepared by mixing 1.00 g of ethanol, C2H5OH, with 100.0 g water. Calculate the mass percent of ethanol in this solution A 135 g sample of seawater is evaporated to dryness, leaving 4.73 g of solid residue (the salts formerly dissolved in the seawater). Calculate the mass percent of solute present in the original seawater

Example: Determine Mass of Solute Although milk is not a true solution (it is really a suspension of tiny globules of fat, protein, and other substrates in water), it does contain a dissolved sugar called lactose. Cow’s milk typically contains 4.5 % by mass of lactose, C12H22O11. Calculate the mass of lactose present in 175 g of milk

15.4 Solution Composition Molarity: the number moles of solute per volume of solution in liters moles of solute Molarity = ------------------------ Liters of solution unit = moles/L or M (molar) Standard solution: is a solution whose concentration is accurately known.

Examples Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution Calculate the molarity of a solution prepared by dissolving 1.56 g of gaseous HCl into enough water to make 26.8 mL of solution Determine how much volume (in ml) will be needed to dissolved 2.50 g of solid NaCl to make 0.050M solution.

E.g Solution Composition: Calculating Ion Concentration Give the concentration of all the ions in each of the following solutions: 0.50 M Co(NO3)2 1.0 M FeCl3

E.g Solution Composition: Calculating Number moles from Molarity How many moles of Ag+ ions are present in 25.0 mL of a 0.75 M AgNO3 solution? How many moles of Na+ ions are present in 42.0 mL of 0.350M NaCl?

Examples: Calculating mass from molarity To analyze the alcohol content of a certain wine, a chemist needs 1.00L of an aqeuous 0.200 M K2Cr2O7 (potassium dichromate) solution. How much solid K2Cr2O7 (molar mass = 294.2 g) must be weighed out to make this solution? Formalin is an aqueous solutions of formaldehyde, HCHO,, used as a preservative for biological speciments. How many grams of formaldehyde must be used to prepare 2.5 L of 12.3 M formalin?

15.5 Dilution Reducing the original concentration of a chemical solution A process of transferring solution to achieve a the desired molarity by diluting with solvent Moles of solute after dilution = moles of solute before dilution Formula  M1 V1 = M2 V2

Examples What volume of 16 M sulfuric acid must be used to prepare 1.5L of 0.10 M of H2SO4 solution? Calculate the new molarity if a dilution is made for: 25.0 ml of water is added to 10.0 mL of 0.251 M CaCl2

15.6 Stoichiometry of Solution Reactions Steps for solving stoichiometric problems involving solutions Step 1: Write a balanced equation for the reaction. For each reactions involving ions, it is best to write the net ionic equation. Step 2: Calculate the moles of reactant Step 3: Determine which reactant is limiting Step 4: Calculate the moles of other reactants or products, as required Step 5: Convert to grams or other units, if required

Examples When Ba(NO3)2 and K2CrO4 react in aqueous solution, the yellow solid BaCrO4 is formed. Calculate the mass of BaCrO4 that forms when 3.50 x 10-3 mole of solid Ba(NO3)2 is dissolved in 265 mL of 0.0100 M K2CrO4 solution

Examples When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed

Neutralization Reaction Use stoichiometry to determine how much of acid or base must be used to reach neutralization Strong acid: HCl(aq)  H+(aq) + Cl-(aq) Strong base: NaOH(s)  Na+(aq) + -OH(aq) Net equation: H+(aq) + -OH(aq)  H2O(l)

Example What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of a 0.350 M NaOH? Calculate the volume of 0.10 M HNO3 needed to neutralize 125 mL of 0.050 MKOH

15.8 Solution Composition: Normality Normality is another unit of concentration sometime used when dealing with acid and base H+ and –OH Equivalent of an acid: the mount of acid that can be furnish 1 mol of H+ ions Equivalent of a base: the amount of that base that can furnish 1 mol of –OH ions

Equivalent E.g 1 mol HCl = 1 equiv HCl molar mass (HCl) = equivalent weight (HCl) 1 mol KOH = 1 equiv KOH molar mass KOH = 1 equiv KOH ½ mol H2SO4 = 1 equiv H2SO4 ½ molar mass H2SO4 = 1 equiv H2SO4

Solution Stoichiometry: Calculating Equivalent Weight Phosphoric acid, H3PO4 can furnish three H+ ions per molecule. Calculate the equivalent weight of H3PO4. Calculate the equivalent weight of HBr

Normality (N) Normality (N) = number of equivalent of solute per liter of solution Knowing Normality can help us calculate The number of equivalents The volume of solution

Calculating Normality A solution of sulfuric acid contains 86 g of H2SO4 per liter of solution. Calculate the normality of this solution Calculate the normality of a solution containing 23.6 g of KOH in 755 ml of solution

Neutralization The number of H+ ions furnished by the sample of acid is the same as the number of –OH ions furnished by the sample of base reacts exactly with n equiv acid --------------- n equiv base Nacid x Vacid = Nbase x Vbase