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A solution is a homogeneous mixture of two or more substances.

A solution is a homogeneous mixture of two or more substances. A solution forms when a substance dissolves at the atomic level in a solvent. If the substance being dissolved is ionic, the atomic level is ions. If the substance being dissolved is covalent, the atomic level is molecules.

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A solution is a homogeneous mixture of two or more substances.

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  1. A solution is a homogeneous mixture of two or more substances. A solution forms when a substance dissolves at the atomic level in a solvent. If the substance being dissolved is ionic, the atomic level is ions. If the substance being dissolved is covalent, the atomic level is molecules. If the substance being dissolved is atomic (like a noble gas), the atomic level is atoms. This is in contrast to a heterogeneous mixture where the solute will not break down to atomic level. I.E. the particles are larger than ions, atoms or molecules. Examples like this include suspensions and colloids.

  2. A solution is a homogeneous mixture of two or more substances. A solvent is a substance that is present in the largest amount in a solution. The solvent dissolves the other compound(s). A solute is a substance that is present in a lesser amount in a solution. The solute dissolves in the solvent. If a substance will dissolve in a solvent, then the substance is soluble. If a substance will not dissolve in a solvent, then the substance is insoluble. If a liquid substance will dissolve in a liquid solvent, the two substances are miscible. If a liquid substance will not dissolve in a liquid solvent, the two substances are immiscible.

  3. Solution formation (solid dissolving in water) When the ions become surrounded by water molecules, they are said to be solvated. Solvation is the process of surrounding a solute particle with solvent molecules. If the solvent is water, the process can also be called hydration. If the intermolecular forces between the water molecules and the ions are stronger than the forces between the ions themselves, the ionic substance will dissolve.

  4. An aqueous solution of ions will conduct electricity. a) Solution of a strong electrolyte like salt b) Solution of a weak electrolyte like a weak acid c) Solution of a non-electrolyte like sugar

  5. NaCl (a salt) breaks apart completely into ions when it dissolves in water. Therefore it is a strong electrolyte. NaCl (s) + H2O (l) → Na+1 (aq) + Cl-1 (aq) Any ionic compound that dissolves in water will produce solvated ions (aq), so all water soluble ionic compounds are strong electrolytes. If a soluble ionic compound contains the OH-1 ion, the ionic compound will be a strong base.

  6. An acid is a substance that breaks apart (dissociates) in water to produce aqueous H+1 ions. If the acid dissociates nearly 100% it is a strong electrolyte. If an acid is a strong electrolyte, it is a strong acid. HCl (a strong acid) breaks apart completely into ions when it dissolves in water. Therefore it is a strong electrolyte. This is also the definition of a strong acid. HCl (g) + H2O (l) → H+1 (aq) + Cl-1 (aq) There are six acids that you need to recognize as strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4

  7. If an acid only dissociates to a small extent (usually less than 5% ions produced), the acid would be classified as a weak acid and would also be a weak electrolyte. Any Acid that is not one of the six strong acids will be a weak acid. When H+1 ions are added to many of the polyatomic anions, the compound you make will usually be an acid. Common examples of weak acids are: HF, CH3CO2H, H3PO4, HClO, HClO2, HClO3 A substance that produces OH-1 ions in water is a base. If a base only produces a few ions (5% or less), the base would be classified as a weak base and would also be a weak electrolyte. Common examples of weak bases are: NH3, Ionic compounds that contain OH-1 but are only slightly soluble

  8. Substances that do not form ions when dissolved in water ate called Nonelectrolytes. Nonelectrolytes are covalent molecules that are soluble in water. Common examples of nonelectrolytes are: Glucose (C6H12O6), Sucrose (C12H22O11), and ethanol (CH3CH2OH) C6H12O6 (s) + H2O (l) → C6H12O6 (aq)

  9. Concentration: describing how much solute is dissolved in a solvent A concentrated solution has a lot of solute dissolved. While a dilute solution has only a little solute dissolved. There are five ways that we will quantify concentration in this class. (mass of solute/mass of solution)X100 Percent by Mass Percent by Volume (volume of solute/volume of solution)X100 Molarity (moles of solute/liter of solution) Molality (moles of solute/kilogram of solvent) (moles of solute)/(moles of solute + moles of solvent) Mole Fraction

  10. ( ) ( ) 1 mol C6H12O6 5.10 g C6H12O6 180.0 g C6H12O6 ( ) ( ) 1 L 100.5 mL 1000 mL (moles of solute)/(L of solution) Molarity (M): Units are: mole/L If the amount of solute is given in grams, you need to convert from grams to moles first! Example Problem: An intravenous solution contains 5.10 g of glucose (C6H12O6) in 100.5 mL of solution. What is the molarity of the solution? Volume of solution: 100.5 mL Mass of Solute: 5.10 g C6H12O6 Molar Mass of C6H12O6: 180.0 g/mol Divide the moles by the L to get molarity (M)

  11. Preparing solutions of a specific concentration: Calculations Steps: Step 1: identify the volume of solution to be made. Step 2: identify the concentration of solution to be made. Step 3: Use the concentration and volume information to determine the “amount” of solute needed. Step 4: If the amount is in moles, convert it to grams using molar mass. Step 5: If the solute is a liquid, convert the mass into volume using density.

  12. Examples: a) How many grams of methanol (CH3OH) would be needed to make 500.0 mL of a 1.50 molar solution? Volume = 500.0 mL = 0.5000 L Concentration: 1.50 molar (M) = 1.50 mol/L n = (1.50 mol/L)(0.5000 L) = 0.750 mol CH3OH So: n = MV Grams CH3OH = (0.750 mol CH3OH)(32.04 g/mol) = 24.0 g CH3OH b) How many mL of CH3OH would be needed for this solution if the density of CH3OH is 0.783 g/mL? So: volume = mass/density density = mass/volume volume = (24.0 g)/0.783 g/mL) = 30.7 mL

  13. Steps Involved in the Preparation of a Standard Aqueous Solution using a solid compound. • Measure the mass of the correct number of grams of solid compound. • Add some water and shake until all solid is dissolved. • Carefully add enough water to fill the container to the volume marker.

  14. Steps Involved in the Preparation of a Standard Aqueous Solution using a liquid compound. • Measure out the correct volume of liquid compound using a pipette or burette. • Add some water and shake until all liquid is dissolved. • Carefully add enough water to fill the container to the volume marker.

  15. Dilution Problems: If you increase the amount of solvent in a solution without changing the amount of solute, you are doing a dilution problem. In a dilution problem the number of moles of solute stays the same while the volume and the concentration of the solution change. Remember: So: n = MV Now in a dilution problem, there are two sets of conditions, the initial and the final (1 and 2 respectively). M1V1 = mol of solute and M2V2 = mol of solute so n1 = n2 and M1V1 = M2V2 Use this form of the equation to solve dilution problems!

  16. Five Types of Chemical Reactions: (SOL level of definition) Synthesis: Two or more reactants combine to make a more complicated product. AB + X → ABX Combustion: Any rapid reaction with oxygen (O2 gas) to produce heat and/or light. AB + O2 → AO + BO Decomposition: One complicated reactant breaking down to make two or more products. ABX → AB + X Double Replacement: AX + BY → AY + BX Single Replacement: AX + Y → AY + X

  17. Double Replacement: AX + BY → AY + BX The reactants arealways aqueous ionic compounds. One of the products must be a solid, a liquid (usually water), or a gas while the other product will be a new aqueous ionic compound. A reaction that makes a solid: (these are also called precipitation reactions) Na2SO4 (aq) + BaCl2 (aq) → 2 NaCl (aq) + BaSO4 (s) A reaction that makes water: (these are also called acid-base reactions) H2SO4 (aq) + Sr(OH)2 (aq) → 2 H2O (l) + SrSO4 (aq) A reaction that makes a gas: Na2S (aq) + 2 HCl (aq) → H2S (g) + 2 NaCl (aq)

  18. For double replacement reactions that produce precipitates, switch ions and write correct ionic formulas for the products-then balance the equation. For Example Mix: Al2(SO4)3 (aq) + BaCl2 (aq) Take the ionic compounds apart: Ba+2 ions Cl-1 ions Al+3 ions SO4-2 ions Switch the ions around-remember positive always mixes with negative: Al+3 ions Ba+2 ions Cl-1 ions SO4-2 ions Write correct formulas for ionic compounds-remember the net charge is 0: AlCl3 + BaSO4 These are the possible products Write a skeleton equation: Al2(SO4)3 (aq) + BaCl2 (aq) → AlCl3 (aq) + BaSO4 (s) Balance to get the chemical equation (also called the molecular equation): Al2(SO4)3 (aq) + 3 BaCl2 (aq) → 2 AlCl3 (aq) + 3 BaSO4 (s)

  19. Illustration of a precipitation reaction between Ba(NO3)2 (aq) and K2CrO4 (aq) a) Ions present when the two solutions are first mixed. b) Ions remaining after formation of precipitate (shown on the bottom of the beaker). K+1 and NO3-1 are spectator ions The reaction in chemical equation form: Ba(NO3)2 (aq) + K2CrO4 (aq) → BaCrO4 (s) + 2 KNO3 (aq)

  20. Chemical Equation or Molecular Equation (balanced) Ba(NO3)2 (aq) + K2CrO4 (aq) → BaCrO4 (s) + 2 KNO3 (aq) Complete Ionic Equation (must show all aqueous ions separated) (reactant side) Ba+2 (aq) + 2 NO3-1 (aq) + 2 K+1(aq) + CrO4-2 (aq) → → BaCrO4 (s) + 2 K+1 (aq) + 2 NO3-1 (aq) (product side) Net Ionic Equation (spectator ions are removed) Ba+2 (aq) + CrO4-2 (aq) → BaCrO4 (s) K+1 (aq) and NO3-1 (aq) Spectator ions: Note: the aqueous ionic product always contains the Spectator ions!

  21. Spectator ions are the ions that remain in the solution after the reaction. The spectator ions do not take part in the reaction. The spectator ions remain in solution after the precipitate forms and the ions in the precipitate are removed from the solution (settle out on the bottom)

  22. KCl (aq) + AgNO3 (aq) → AgCl (s) + KNO3 (aq) 1st solution 2nd solution mixing Final solution K+1 (aq) + Cl-1 (aq) + Ag+1 (aq) + NO3-1(aq) → AgCl (s) + K+1 (aq) + NO3-1 (aq) Can you identify the spectator ions?

  23. There are three types of double replacement reactions: • Those that form a solid ionic precipitate. • Those that form water. • Those that produce a gas. These are also called precipitation reactions. These are also called acid-base reactions. These have no other specific name. If switching the ions in a double replacement reaction does not do one or more of these three things, then no reaction is taking place.

  24. Double replacement reactions that produce water (acid-base) Hydrobromic acid plus sodium hydroxide HBr (aq) + NaOH (aq) → H2O (l) + NaBr (aq) H+1 (aq) + Br-1 (aq) + Na+1 (aq) + OH-1 (aq) → H2O (l) + Na+1 (aq) + Br-1 (aq) H+1 (aq) + OH-1 (aq) → H2O (l) The key to identifying these reactions is to recognize acids and bases. Adding H+1 ions to most of the negative ions we have used so far will produce an acid. The only bases we have at this point are OH-1 and CO3-2 Therefore, hydroxide compounds and carbonate compounds are bases.

  25. Double replacement reactions that produce a gas. 2 HI (aq) + Li2S (aq) → H2S (g) + 2 LiI (aq) 2 H+1 (aq) + 2 I-1 (aq) + 2 Li+1 (aq) + S-2 (aq) → H2S (g) + 2 Li+1 (aq) + 2 I-1 (aq) 2 H+1 (aq) + S-2 (aq) → H2S (g) 2 HCl (aq) + Na2CO3 (aq) → H2CO3 (aq) + 2 NaCl (aq) But carbonic acid is not stable and falls apart! H2CO3 (aq) → H2O (l) + CO2 (g) Therefore: 2 HCl (aq) + Na2CO3 (aq) → H2O (l) + CO2 (g) + 2NaCl (aq) The are not very many of these reactions. Common ones produce H2S or CO2.

  26. For each of the following: Identify the type of reaction, identify the spectator ions, and write the net ionic equation. You may want to write the complete ionic equation to help you identify the spectator ions. AgNO3 (aq) + CaBr2 (aq) → AgBr (s) + Ca(NO3) 2 (aq) Spectators: Ca+2 and NO3-1, double replacement: precipitation KOH (aq) + HI (aq) → KI (aq) + H2O (l) Spectators: K+1 and I-1, double replacement: acid-base HNO3 (aq) + Ca(OH)2 (aq) → H2O (l) + Ca(NO3) 2 (aq) Spectators: Ca+2 and NO3-1, double replacement: acid-base HClO4 (aq) + Na2S (aq) → H2S (g) + NaClO4 (aq) Spectators: Na+1 and ClO4-1, double replacement: produces a gas Pb(NO3)2 (aq) + Li2SO4 (aq) → LiNO3 (aq) + PbSO4 (s) Spectators: Li+1 and NO3-1, double replacement: precipitation

  27. Rules for predicting precipitates: Slightly soluble means that they will be a precipitate. If a compound contains ions other than those mentioned, assume they are insoluble (form a precipitate) unless told otherwise.

  28. Practice: Use the rules on the previous page to predict whether the following ions will be soluble in water. If they are not soluble, then they would be precipitates in double replacement reactions. NaCl Na2CO3 K2CO3 AgCl soluble soluble soluble insoluble AgNO3 CaBr2 AgBr BaSO4 soluble soluble insoluble insoluble Fe2(CO3)3 Li2S Al(OH)3 (NH4)3PO4 insoluble soluble insoluble soluble Mg3(PO4)2 PbBr2 AgBr MgSO4 insoluble insoluble insoluble soluble

  29. Synthesis: Two or more reactants combine to make a more complicated product. AB + X → ABX CaO (s) + H2O (l) → Ca(OH)2 (s) The reactants may be compounds (ionic or covalent) or may be elements. The product must always be a compound (ionic or covalent). If more than one kind of product is made, the reaction is not technically a synthesis reaction.

  30. Decomposition: One complicated reactant breaking down to make two or more products. ABX → AB + X Fe2(CO3) 3 (s) → Fe2O3 (s) + 3 CO2 (g) The reactant must always be compounds (ionic or covalent). The products may be compounds (ionic or covalent) or elements.

  31. Single Replacement: AX + Y → AY + X or A + BX → AX + B An ionic compound is reacting with an element. The element (charge of zero) will become an ion and replace an ion in the ionic compound. The ion that was replaced will become an element (charge of zero) SnBr2 (s) + Fe (s) → FeBr3 (s) + Sn (s) The element iron (Fe) is replacing the Sn+2 ion. When it does this, iron becomes an ion (Fe+3) while the Sn+2 ion becomes an element (Sn).

  32. Activity Series for Single Replacement Reactions Metals Most Reactive Halides (halogens) Li Rb K Ca Na Mg Al Mn Zn Fe Ni Sn Pb Cu Ag Pt Au Most Reactive F Cl Br I Least Reactive Only a more reactive element can replace another element in a compound! The element produced must be less reactive than the starting element! Least Reactive

  33. Will the reaction below will work? 2 LiBr (s) + F2 (g) → 2 LiF (s) + Br2 (l) Yes because F is more reactive than Br Will the reaction below will work? 2 LiBr (s) + I2 (g) → 2 LiI (s) + Br2 (l) No because I is less reactive than Br Will the reaction below will work? AgBr (s) + Li (s) → LiBr (s) + Ag (s) Yes because Li is more reactive than Ag Will the reaction below will work? CaBr2 (s) + Zn (s) → ZnBr2 (s) + Ca (s) No because Zn is less reactive than Ca

  34. Notice that Halogen atoms (group 17) always replace halogen ions! 2 LiBr (s) + F2 (g) → 2 LiF (s) + Br2 (l) Here the element fluorine (F2) is replacing the ion Br-1 Notice that Metal atoms always replace metal ions! AgBr (s) + Li (s) → LiBr (s) + Ag (s) Here the element lithium (Li) is replacing the ion Ag+1 Notice that the ion that was replaced always becomes an element!

  35. Pop Quiz: • Write the complete ionic equation for the following reaction. • b) Write the the net ionic equation for the following reaction. • c) Identify any spectator ions. Group A 3 Ca(NO3)2 (aq) + Au2(SO4)3 (aq) → 3 CaSO4 (s) + 2 Au(NO3)3 (aq) Group B 2 AlCl3 (aq) + 3 Na2CO2 (aq) → 6 NaCl (aq) + Al2(CO2)3 (s)

  36. Acid-Base Titration Titration is a process where the concentration of an acidic solution can be determined by reacting the acid with a measured volume of a basic solution of known concentration. Titrations make use of indicators. An indicator is a substance that changes color when the pH of the solution changes. A common indicator is Phenolphthalein. Colorless in acid and Pink in base.

  37. The point during a titration where the moles of base added is equal to the moles of acid originally present is called the equivalence point. In order for an indicator to be useful for a titration, the color change must occur at or very near to the equivalence point. For strong acid-strong base titrations, the equivalence point will be at pH of 7. So indicators that change color at or very near pH 7 will work.

  38. The Useful pH Ranges for Several Common Indicators

  39. Why is Phenolphthalein a good indicator even though its color change occurs at a pH of about 8.5? One: Its acidic form is colorless, so that makes the color change easier to detect. Two: The titration curve is nearly vertical near the equivalence point, so adding one or two drops of base will cause a very large change in pH. The error associated with one or two drops of added base is small. Calculations for the titration of 50.0 mL of 0.20 M HCl with 0.200 M NaOH: At the start: pH = 0.70 and mole HCl = (0.200 mol/L)(0.0500 L) = 0.0100 mol After 10.0 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0100 L) = 0.00200 mol Mole of HCl remaining = (0.0100 mol – 0.00200 mol) = 0.0080 mol Concentration of HCl now = (0.0080 mol/(0.0500 L + 0.0100 L)) = 0.13 M pH now = -log(0.13) = 0.89

  40. After 30.0 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0300 L) = 0.00600 mol Mole of HCl remaining = (0.0100 mol – 0.00600 mol) = 0.0040 mol Concentration of HCl now = (0.0040 mol/(0.0500 L + 0.0300 L)) = 0.050 M pH now = -log(0.050) = 1.30 After 49.5 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0495 L) = 0.00990 mol Mole of HCl remaining = (0.0100 mol – 0.00990 mol) = 0.0001 mol Concentration of HCl now = (0.0001 mol/(0.0500 L + 0.0495 L)) = 0.001 M pH now = -log(0.001) = 3.0

  41. After 50.5 mL of NaOH has been added: Mole NaOH added = (0.200 mol/L)(0.0505 L) = 0.0101 mol Mole of HCl remaining = (0.0100 mol – 0.0101 mol) = no meaning for negative numbers Concentration of NaOH now = (0.0101 mol - 0.0100 mol) = 0.0001 mol = (0.0001 mol/(0.0500 L + 0.0505 L) = 0.001 mol/L pOH now = -log(0.001) = 3.0 pH = 14.0 – 3.0 = 11.0

  42. A titration curve: an acid being titrated with NaOH.

  43. The pH Curve for the Titration of 100.0 mL of 0.10 M of HCl with 0.10 M NaOH

  44. The pH Curve for the Titration of 50 mL of 0.1 M HC2H3O2 with 0.1 M NaOH; Phenolphthalein Will Give an End Point Very Close to the Equivalence Point of the Titration

  45. Weak Conjugate Acid Weak Conjugate Base Summary for predicting pH of solutions at the equivalence point of a titration. pH at Equivalence Point Strong Acid + Strong Base 7.00 Strong Acid + Weak Base Acidic Weak Acid + Strong Base Basic Actual pH depends upon Ka and Kb values of the conjugates.

  46. Redox: Oxidation-Reduction Reactions An Oxidation-Reduction Reaction is any reaction where one substance gains electrons while another substance loses electrons. 2 Na (s) + Cl2 (g) → 2 Na+ + 2 Cl- (ions in the crystal) What happened to Na? What happened to Cl?

  47. Na (s) → Na+ + e- Each Na is losing an electron to become a cation Oxidation is losing electrons Cl2+ 2 e- → 2 Cl- Each Cl is gaining an electron to become an anion Reduction is gaining electrons OILRIG Oxidation Is Loss Reduction Is Gain

  48. Since metals have low electronegativities, they tend to lose electrons (be oxidized) while nonmetals which have higher electronegativities tend to gain electrons (be reduced) A substance that is losing electrons is becoming more positive. This means that the oxidation state is increasing. A substance that is gaining electrons is becoming more negative. This means that the oxidation state is decreasing. An agent is a substance that causes something to happen. If it causes an oxidation to occur, it would be an oxidizing agent. If it causes a reduction to occur, it would be a reducing agent.

  49. 2 NaBr (aq) + Cl2 (aq) → 2 NaCl (aq) + Br2 What substance is being oxidized? What substance is being reduced? What substance is the oxidizing agent? What substance is the reducing agent? How many electrons are transferred in the balanced reaction? Oxidation half reaction 2 Br-→ Br2 + 2 e- Cl2 + 2 e-→ 2 Cl- Reduction half reaction Na+→ Na+ (spectator ion) Since Br- is oxidized, Cl2 must have caused it to happen, so Cl2 is the oxidizing agent. Since Cl2 is reduced, Br- must have caused it to happen, so Br- is the reducing agent. Oxidation States or Oxidation Numbers are helpful in identifying what is being oxidized or reduced, and they let you determine the number of electrons gained or lost.

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