Solution Chemistry (Chp. 7)

Solutions Terms Molar Concentration (mol/L) Dilutions % Concentration (pp. 255 – 263) Solution Process Solution Preparation Solution Stoichiometry Dissociation

Terms solution solvent solute concentrated dilute aqueous miscible immiscible alloy solubility molar solubility saturated unsaturated supersaturated dissociation electrolyte non-electrolyte soluble insoluble limiting reagent excess reagent actual yield theoretical yield decanting pipetting filtrate precipitate dynamic equilibrium

Define the terms in bold and italics from pp. 237 – 240. Solids, liquids, and gases can combine to produce 9 different types of solution. Give an example of each type. p. 242 #’s 5, 7, 9, & 10

Terms solution - is a homogeneous mixture solute - the substance that dissolves OR the substance in lesser quantity solvent - the substance which dissolves the solute OR the substance in greater quantity concentrated - a large amount of solute relative to the amount of solvent dilute - a small amount of solute relative to the amount of solvent

Terms saturated – contains the maximum amount of dissolved solute at a given temperature and pressure unsaturated – contains less than the maximum amount of dissolved solute at a given temperature and pressure supersaturated – contains more than the maximum amount of dissolved solute for a given temperature and pressure

Terms miscible – liquids that dissolve in each other immiscible – liquids that do not dissolve in each other aqueous - the solvent is water alloy - a solid solution of two or more metals

Terms Solubility - the maximum amount of solute that can be dissolved under specific temperature and pressure conditions eg. the solubility of HCl at 25 °C is 12.4 mol/L eg. 100.0 mL of water at 25°C dissolves 36.2 g of sodium chloride

Terms soluble – solubility is greater than 1 g per 100 mL of solvent. insoluble - solubility is less than 0.1 g per 100 mL of solvent.

9 Types of Solution

Factors Affecting Solubility (pp.243 – 254) List 3 factors that affect the rate of dissolving. How does each of the following affect solubility? particle size temperature pressure

Rate of Dissolving for most solids, the rate of dissolving is greater at higher temperatures stirring a mixture or by shaking the container increases the rate of dissolving. decreasing the size of the particles increases the rate of dissolving.

Solubility small molecules are often more soluble than larger molecules. solubility of solids increases with temperature. the solubility of most liquids is not affected by temperature. the solubility of gases decreases as temperature increases an increase in pressure increases the solubility of a gas in a liquid.

Factors Affecting Solubility What type of solvent will dissolve: polar solutes and ionic solutes nonpolar solutes Why do some ionic compounds have low solubility in water? p. 254 #’s 1, 2, 4 - 6

“Like Dissolves Like” ionic solutes and polar covalent solutes both dissolve in polar solvents non-polar solutes dissolve in non-polar solvents. compounds with very strong ionic bonds, such as AgCl, tend to be less soluble in water than compounds with weak ionic bonds, such as NaCl.

Applications 1. An opened soft drink goes ‘flat’ faster if not refrigerated. 2. Thermal pollution (warming lake water) is not healthy for the fish living in it. 3. After pouring 5 glasses of pop from a 2 litre container, Jonny stoppered the bottle and crushed it to prevent the remaining pop from going flat.

Molar Concentration Review: - Find the molar mass of Ca(OH)2 - How many moles in 45.67 g of Ca(OH)2? - Find the mass of 0.987 mol of Ca(OH)2.

Molar Concentration The terms concentrated and dilute are qualitative descriptions of solubility. A quantitative measure of solubility uses numbers to describe how much solute is dissolved or the concentration of a solution.

Molar Concentration The MOLAR CONCENTRATION of a solution is the number of moles of solute (n) per litre of solution (v).

V Molar Concentration FORMULA: Molar Concentration = number of moles volume in litres C = n V

eg. Calculate the molar concentration of: 4.65 mol of NaOH is dissolved to prepare 2.83 L of solution. (1.64 mol/L) 15.50 g of NaOH is dissolved to prepare 475 mL of solution. ( 0.3875 mol → 0.816 mol/L) p. 268 - # 19

eg. Calculate the following: Rearranged Formulas eg. Calculate the following: the number of moles in 4.68 L of 0.100 mol/L KCl solution. (0.468 mol) the mass of KCl in 268 mL of 2.50 mol/L KCl solution. (0.670 mol → 49.9 g) p. 268 #’s 20 - 24

the volume of 6. 00 mol/L HCl(aq) that can be made using 0 the volume of 6.00 mol/L HCl(aq) that can be made using 0.500 mol of HCl. the volume of 1.60 mol/L HCl(aq) that can be made using 20.0 g of HCl.

Dilution (p. 272) The number of moles remains the same Number of moles before dilution When a solution is diluted: The concentration decreases The volume increases The number of moles remains the same ni = nf Number of moles after dilution

Ci Vi = Cf Vf Dilution (p. 272) ni = nf eg. Calculate the molar concentration of a vinegar solution prepared by diluting 10.0 mL of a 17.4 mol/L solution to a final volume of 3.50 L.

p. 273 #’s 25 – 27 p. 276 #’s 1, 2, 4, & 5 DON’T SHOW UP UNLESS THIS IS DONE!!

Solution Preparation & Dilution standard solution – a solution of known concentration volumetric flask – a flat-bottomed glass vessel that is used to prepare a standard solution delivery pipet – pipets that accurately measure one volume graduated pipet – pipets that have a series of lines that can be use to measure many different volumes

To prepare a standard solution: 1. calculate the mass of solute needed 2. weigh out the desired mass 3. dissolve the solute in a beaker using less than the desired volume 4. transfer the solution to a volumetric flask (rinse the beaker into the flask) 5. add water until the bottom of the meniscus is at the etched line

To dilute a standard solution: 1. Rinse the pipet several times with deionized water. 2. Rinse the pipet twice with the standard solution. 3. Use the pipet to transfer the required volume. 4. Add enough water to bring the solution to its final volume.

Percent Concentration Concentration may also be given as a %. The amount of solute is a percentage of the total volume/mass of solution. liquids in liquids - % v/v solids in liquids - % m/v solids in solids - % m/m

Percent Concentration p. 258 #’s 1 – 3 DSUUTID!!

p. 261 #’s 5 – 9 DSUUTID!!

p. 263 #’s 10 – 13 DSUUTID!!

Concentration in ppm and ppb Parts per million (ppm) and parts per billion (ppb) are used for extremely small concentrations

eg. 5. 00 mg of NaF is dissolved in 100. 0 kg of solution eg. 5.00 mg of NaF is dissolved in 100.0 kg of solution. Calculate the concentration in: a) ppm b) ppb

ppm = 0.005 g x 106 100,000 g = 0.05 ppm ppb = 0.005 g x 109 = 50.0 ppb

DON’T SHOW UP UNLESS THIS IS DONE!! p. 265 #’s 15 – 17 pp. 277, 278 #’s 11, 13, 15 – 18, 20 DON’T SHOW UP UNLESS THIS IS DONE!!

Solution Stoichiometry 1. Write a balanced equation 2. Calculate moles given n=m/M OR n=CV 3. Mole ratios 4. Calculate required quantity

Solution Stoichiometry eg. 45.0 mL of a HCl(aq) solution is used to neutralize 30.0 mL of a 2.48 mol/L NaOH solution. Calculate the molar concentration of the HCl(aq) solution. p. 304: #’s 16, 17, & 18 Worksheet

Sample Problems 1. What mass of copper metal is needed to react with 250.0 mL of 0.100 mol/L silver nitrate solution? 2. Calculate the volume of 2.00 M HCl(aq) needed to neutralize 1.20 g of dissolved NaOH. 3. What volume of 3.00 mol/L HNO3(aq) is needed to neutralize 450.0 mL of 0.100 mol/L Sr(OH)2(aq)?

Sample Problem Solutions Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq) Step 2 n = 0.02500 mol AgNO3 Step 3 n = 0.01250 mol Cu Step 4 m = 0.794 g Cu

Sample Problem Solutions HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Step 2 n = 0.0300 mol NaOH Step 3 n = 0.0300 mol HCl Step 4 V = 0.0150 L HCl

Sample Problem Solutions 2 HNO3(aq) + Sr(OH)2(aq) → 2 H2O(l) + Sr(NO3)2(aq) Step 2 n = 0.04500 mol Sr(OH)2 Step 3 n = 0.0900 mol HNO3 Step 4 V = 0.0300 mol/L HNO3

The Solution Process (p. 299) Dissociation occurs when an ionic compound breaks into ions as it dissolves in water. A dissociation equation shows what happens to an ionic compound in water. eg. NaCl(s) → Na+(aq) + Cl-(aq) K2SO4(s) → 2 K+(aq) + SO42-(aq) Ca(NO3)2(s) → Ca2+(aq) + 2 NO3-(aq)

The Solution Process (p. 299) Solutions of ionic compounds conduct electric current. A solute that conducts an electric current in an aqueous solution is called an electrolyte.

The Solution Process (p. 299) Acids are also electrolytes. Acids form ions when dissolved in water. eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq) HCl(s) → H+(aq) + Cl-(aq)

The Solution Process (p. 299) Molecular Compounds DO NOT dissociate in water. eg. C12H22O11(s) → C12H22O11(aq) Because they DO NOT conduct electric current in solution, molecular compounds are non-electrolytes.

The Solution Process (p. 299) The molar concentration of any dissolved ion is calculated using the ratio from the dissociation equation. eq. What is the molar concentration of each ion in a 5.00 mol/L MgCl2(aq) solution: 5.00 mol/L 5.00 mol/L 10.00 mol/L

p. 300 #’s 7 – 9 What mass of calcium chloride is required to prepare 2.00 L of 0.120 mol/L Cl-(aq) solution? p. 302 # 14 p. 311 #’s 11, 12, 16, & 18