1 Chapter 4 Aqueous solutions Types of reactions

2 Parts of Solutions l Solution- homogeneous mixture. l Solute- what gets dissolved. l Solvent- what does the dissolving. l Soluble- Can be dissolved. l Miscible- liquids dissolve in each other.

3 Aqueous solutions l Dissolved in water. l Water is a good solvent because the molecules are polar. l The oxygen atoms have a partial negative charge. l The hydrogen atoms have a partial positive charge. l The angle is 105ºC.

4 Hydration l The process of breaking the ions of salts apart. l Ions have charges and attract the opposite charges on the water molecules.

5 Hydration H H O H H O H H O H H O H H O H H O H H O H H O H H O

6 Solubility l How much of a substance will dissolve in a given amount of water. l Usually g/100 mL l Varies greatly, but if they do dissolve the ions are separated, l and they can move around. l Water can also dissolve non-ionic compounds if they have polar bonds.

7 Electrolytes l Electricity is moving charges. l The ions that are dissolved can move. l Solutions of ionic compounds can conduct electricity. l Electrolytes. l Solutions are classified three ways.

8 Types of solutions l Strong electrolytes- completely dissociate (fall apart into ions). l Many ions- Conduct well. l Weak electrolytes- Partially fall apart into ions. l Few ions -Conduct electricity slightly. l Non-electrolytes- Don’t fall apart. l No ions- Don’t conduct.

9 Types of solutions l Acids- form H + ions when dissolved. l Strong acids fall apart completely. l many ions l H 2 SO 4 HNO 3 HCl HBr HI HClO 4 l Weak acids- don’ dissociate completely. l Bases - form OH - ions when dissolved. l Strong bases- many ions. l KOH NaOH

10 Measuring Solutions l Concentration- how much is dissolved. l Molarity = Moles of solute Liters of solution l abbreviated M l 1 M = 1 mol solute / 1 liter solution l Calculate the molarity of a solution with 34.6 g of NaCl dissolved in 125 mL of solution.

11 Molarity l How many grams of HCl would be required to make 50.0 mL of a 2.7 M solution? l What would the concentration be if you used 27g of CaCl 2 to make 500. mL of solution? l What is the concentration of each ion?

12 Molarity l Calculate the concentration of a solution made by dissolving 45.6 g of Fe 2 (SO 4 ) 3 to 475 mL. l What is the concentration of each ion?

13 Making solutions l Describe how to make mL of a 1.0 M K 2 Cr 2 O 4 solution. l Describe how to make 250. mL of an 2.0 M copper (II) sulfate dihydrate solution.

14 Dilution l Adding more solvent to a known solution. l The moles of solute stay the same. l moles = M x L l M 1 V 1 = M 2 V 2 l moles = moles l Stock solution is a solution of known concentration used to make more dilute solutions

15 Dilution l What volume of a 1.7 M solutions is needed to make 250 mL of a 0.50 M solution? l 18.5 mL of 2.3 M HCl is added to 250 mL of water. What is the concentration of the solution? l 18.5 mL of 2.3 M HCl is diluted to 250 mL with water. What is the concentration of the solution?

16 Dilution l You have a 4.0 M stock solution. Describe how to make 1.0L of a.75 M solution. l 25 mL 0.67 M of H 2 SO 4 is added to 35 mL of 0.40 M CaCl 2. What mass CaSO 4 Is formed?

17 Types of Reactions  Precipitation reactions l When aqueous solutions of ionic compounds are poured together a solid forms. l A solid that forms from mixed solutions is a precipitate l If you’re not a part of the solution, your part of the precipitate

18 Precipitation reactions NaOH(aq) + FeCl 3 (aq)   NaCl(aq) + Fe(OH) 3 (s) l is really Na + (aq)+OH - (aq) + Fe +3 + Cl - (aq)   Na + (aq) + Cl - (aq) + Fe(OH) 3 (s) l So all that really happens is OH - (aq) + Fe +3  Fe(OH) 3 (s) l Double replacement reaction

19 Precipitation reaction l We can predict the products l Can only be certain by experimenting l The anion and cation switch partners AgNO 3 (aq) + KCl(aq)  Zn(NO 3 ) 2 (aq) + BaCr 2 O 7 (aq)  CdCl 2 (aq) + Na 2 S(aq) 

20 Precipitations Reactions l Only happen if one of the products is insoluble l Otherwise all the ions stay in solution- nothing has happened. l Need to memorize the rules for solubility (pg 145)

21 Solubility Rules  All nitrates are soluble  Alkali metals ions and NH 4 + ions are soluble  Halides are soluble except Ag +, Pb +2, and Hg 2 +2  Most sulfates are soluble, except Pb +2, Ba +2, Hg +2,and Ca +2

22 Solubility Rules  Most hydroxides are slightly soluble (insoluble) except NaOH and KOH  Sulfides, carbonates, chromates, and phosphates are insoluble  Lower number rules supersede so Na 2 S is soluble

23 Three Types of Equations l Molecular Equation- written as whole formulas, not the ions. K 2 CrO 4 (aq) + Ba(NO 3 ) 2 (aq)  l Complete Ionic equation show dissolved electrolytes as the ions. 2K + + CrO Ba NO 3 -  BaCrO 4 (s) + 2K NO 3 - l Spectator ions are those that don’t react.

24 Three Type of Equations l Net Ionic equations show only those ions that react, not the spectator ions Ba +2 + CrO 4 -2  BaCrO 4 (s) l Write the three types of equations for the reactions when these solutions are mixed. l iron (III) sulfate and potassium sulfide Lead (II) nitrate and sulfuric acid.

25 Stoichiometry of Precipitation l Exactly the same, except you may have to figure out what the pieces are. l What mass of solid is formed when mL of M Barium chloride is mixed with mL of M sodium hydroxide? l What volume of M HCl is needed to precipitate the silver from 50.ml of M silver nitrate solution ?

26 Types of Reactions  Acid-Base l For our purposes an acid is a proton donor. l a base is a proton acceptor usually OH - l What is the net ionic equation for the reaction of HCl(aq) and KOH(aq)? Acid + Base  salt + water H + + OH -  H 2 O

27 Acid - Base Reactions l Often called a neutralization reaction Because the acid neutralizes the base. l Often titrate to determine concentrations. l Solution of known concentration (titrant), l is added to the unknown (analyte), l until the equivalence point is reached where enough titrant has been added to neutralize it.

28 Titration l Where the indicator changes color is the endpoint. l Not always at the equivalence point. l A mL sample of aqueous Ca(OH) 2 requires mL of M Nitric acid for neutralization. What is [Ca(OH) 2 ]? l # of H + x M A x V A = # of OH - x M B x V B