Solutions

What are ways that you make a solute dissolve faster? Rate of Solution What are ways that you make a solute dissolve faster? Increase the temperature. Crush or use smaller size solute particles. Stir the solutions.

Types of Solutions There are three ways to classify a solution. Unsaturated Solutions Saturated Solutions Supersaturated Solutions

Relation to Solubility Curve Types of Solutions Type of Solution Description Picture Relation to Solubility Curve Unsaturated Saturated Supersaturated A solution in which more solute can dissolve Below the Line A solution in which contains the maximum amount of solute On the Line See PDF for pictures Above the Line A solution in which contains more than the maximum amount of solute

Supersaturated The solution is holding more solute than it should be able to. This is achieved by heating the solution and then cooling it slowly. Examples: rock candy, southern style sweet tea, chemical heat packs Supersaturated solutions are unstable. The supersaturation is only temporary

SUPERSATURATED SOLUTION Solubility UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration

Supersaturated

Solubility Curves Solubility indicates the amount of solute that will dissolve in a given amount of solvent at a specific temperature. For this curve, X-Axis Temperature Y-Axis How Much Solute Dissolves in 100g of Water Various Lines Each line represents a different solute.

Solubility Curves Supersaturated CO2 Saturated Unsaturated

Using an available solubility curve, classify as unsaturated, saturated, or supersaturated. 80 g NaNO3 @ 30oC per 100 g H2O unsaturated 45 g KCl @ 60oC saturated 30 g KClO3 @ 30oC supersaturated 70 g Pb(NO3)2 @ 60oC unsaturated

Describe each situation below. (A) Per 100 g H2O, 100 g NaNO3 @ 50oC. unsaturated; all solute dissolves; clear solution. (B) Cool solution (A) very slowly to 10oC. supersaturated; extra solute remains in solution; still clear (C) Quench solution (A) in an ice bath to 10oC. saturated; extra solute (20 g) can’t remain in solution and becomes visible

How to use a solubility graph? A. IDENTIFYING A SUBSTANCE ( given the solubility in g/100 cm3 of water and the temperature) Look for the intersection of the solubility and temperature.

Using Solubility Curves How much KNO3 would dissolve in 100g of water at 50oC? How much NH4Cl would dissolve in 200g of water at 70oC? At what temperature would 22g of KCl be able to dissolve in 50g of water? Which is more soluble (has a higher solubility) at 40oC? NH3 KClO3 84g 120g 68oC

Learning Check : What substance has a solubility of 90 g/100 cm3 in water at a temperature of 25ºC ?

Learning Check : What substance has a solubility of 200 g/100 cm3 of water at a temperature of 90ºC ?

Look for the temperature or solubility Locate the solubility curve needed and see for a given temperature, which solubility it lines up with and visa versa.

Learning Check: What is the solubility of potassium nitrate at 80ºC ?

What is the solubility of potassium nitrate at 80ºC ?

Learning Check : At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?

Learning Check: At what temperature will sodium nitrate have a solubility of 95 g/100 cm3 ?

Learning Check: At what temperature will potassium iodide have a solubility of 230 g/100 cm3 ?

Learning Check: At what temperature will potassium iodide have a solubility of 130 g/100 cm3 ?

Using Solubility Curves: What is the solubility of sodium chloride at 25ºC in 100 cm3 of water ? From the solubility graph we see that sodium chlorides solubility is 36 g.

SOLUBLE OR INSOLUBLE? Soluble: able to be dissolved Insoluble: does not dissolve in solution (or water) Precipitate: an insoluble solid formed when two solutions are mixed

Soluble compounds contain… Insoluble compounds contain… For these compounds, common exceptions are INSOLUBLE. SOLUBLE.

Double Replacement reactions AB + CD → AD + CB NaCl + AgNO3 → NaNO3 + AgCl

Concentration of Solute The amount of solute in a solution is given by its concentration. Molarity ( M ) = moles solute liters of solution

Step 1: Calculate moles of NiCl2•6H2O PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough water to make 250 mL of solution. Calculate the Molarity. Known Mass=5 g Volume = 0.250 L Unknown Molarity of NiCl2•6 H2O ? Analysis molar mass = 237.7 g M= n/ V; Step 1: Calculate moles of NiCl2•6H2O Step 2: Calculate Molarity [NiCl2•6 H2O ] = 0.0841 M

MOLARITY PROBLEM What mass of oxalic acid, H2C2O4, is required to make 250. mL of a 0.0500 M solution? Known Volume = 0.250 L M = 0.0500 moles/L Unknown g of H2C2O4,? Analysis molar mass = 90.00 g M= mol/ V; Step 1: Calculate moles of H2C2O4 (0.0500 mol/L) x (0.250 L) = 0.0125 moles Step 2: Convert moles to grams 0.0125 mol H2C2O4 x (90.00 g/mol) = 1.13 g H2C2O4 1 mol H2C2O4

Learning Check How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 1) 12 g 2) 48 g 3) 300 g

Making Molar Solutions …from liquids (More accurately, from stock solutions)

Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) after dilution (f) = MiVi MfVf = 4.5

Making a Dilute Solution remove sample moles of solute initial solution same number of moles of solute in a larger volume mix Making a Dilute Solution diluted solution Timberlake, Chemistry 7th Edition, page 344

Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

Dilution Practice Problem What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? Known Molarity1 = 15.8 mol/L of HNO3 Volume2 = 250 mL Molarity2 = 6.0 mol/L of HNO3 Unknown Volume of HNO3? Equation M1V1 = M2V2 Solution : V1 = (6.0 mol/L x 0.250 L) / 15.8 M = 0.095 L

Dilution Practice Problem If we have 1 L of 3 M HCl, what is M if we dilute acid to 6 L? Known Molarity1 = 3 mol/L Volume1 = 1 L of HCl Volume2 = 6L of HCl Unknown Molarity of HCl? Equation M1V1 = M2V2; Solution : M2 = (3 mol/L x 1 L) / (6 L) = 0.5 M

Dilution Practice Problem What volume of 0.5 M HCl can be prepared from 1 L of 12 M HCl? Solution : V2 = (12 mol/L x 1 L) / (0.5 L) = 24 L Known Volume1 = 1 L of HCl Molarity1 = 12 mol/L of HCl Molarity2 = 0.5 mol/L of HCl Unknown Volume of HCl? Equation M1V1 = M2V2