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Qualitative Analysis:

Qualitative Analysis:. An analysis that determines what’s in a solution, the qualities of the solution. Quantitative Analysis:. An analysis that determines the amounts of a substance present in a solution. Qualitative Analysis. Examples:. 1. Solution Colour. Colorimetry :.

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Qualitative Analysis:

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  1. Qualitative Analysis: An analysis that determines what’s in a solution, the qualitiesof the solution. Quantitative Analysis: An analysis that determines the amounts of a substance present in a solution.

  2. Qualitative Analysis Examples: 1. Solution Colour Colorimetry: An analysis that uses the colour of a solution and the light that passes through it to analyze the solution.

  3. Qualitative Analysis Examples: 1. Solution Colour 2. Flame Test Data Booklet Pg. 6

  4. Quantitative Analysis Examples: 1. Gravimetric Analysis The process of using stoichiometry to calculate masses of unknown substances. Solution Stoichiometry Gas Stoichiometry Gravimetric (solids) Stoichiometry

  5. Quantitative Analysis Examples: 1. Gravimetric Analysis 2. Titration Analysis An analysis that records known volumes and the principles of stoichiometry to calculate concentration, volume, or moles of unknown substance. ***Know your solution stoichiometry really well to do these***

  6. In gravimetric analysis you are using masses and converting them to moles so that you can do stoichiometry. In gravimetric analysis it is often beneficial to mix two solutions together and make a solid precipitate. (solids are easier to weigh and measure). Solid precipitates are produced when ions come together and they form slightly soluble compounds that “fall out of solution”. Example: Predict the products and balance. 1 1 1 1 AgNO3 + NaCl NaNO3 + AgCl

  7. In gravimetric analysis you are using masses and converting them to moles so that you can do stoichiometry. In gravimetric analysis it is often beneficial to mix two solutions together and make a solid precipitate. (solids are easier to weigh and measure). Solid precipitates are produced when ions come together and they form slightly soluble compounds that “fall out of solution”. Example: Now use the solubility chart to determine the states…soluble (aq) or slightly soluble (s) Precipitate 1 1 1 1 AgNO3 + NaCl NaNO3 + AgCl (aq) (aq) (s) (aq)

  8. Precipitate 1 1 1 1 AgNO3 + NaCl NaNO3 + AgCl (aq) (aq) (s) (aq) When your mixing two solutions together you have no idea what the concentration of the two solutions might be. In this situation you want to produce as much solid so you can find out the moles and thus the concentration, but in order to do this you have to make sure ALL the limiting reagent reacts and produces solid product. The extent to which ALL the limiting reagent is reacted by adding more and more excess reagent in a step wise fashion is called Precipitation Completeness. ****You continue to add more and more excess reagent in a drop wise fashion until the mixing of the two liquids NO LONGER produces any “cloudiness”.

  9. In any chemical reaction there is one reagent that will “run out” first….limiting reagent. And one reagent that is present in more amount than can ever be reacted….excess reagent. The question is, how can you tell? Whenever you are asked to calculate excess and limiting reagents you will know the amounts of BOTH reactants.

  10. Example States? 1. Balanced Chemical Equation 2 2 1 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s)

  11. Example 2. Write down what you know. 2 2 1 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s) m = 20.0 g m = 10.0 g

  12. Example 3. Convert all quantities to mols. 2 1 2 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s) m = 20.0 g m = 10.0 g M = 169.88 g/mol M = 63.55 g/mol n = 0.118 mol n = 0.157mol

  13. Example 4. Convert one reagent into the other using N/G and see which is larger/bigger. L.R E.R 2 1 2 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s) m = 20.0 g m = 10.0 g M = 169.88 g/mol M = 63.55 g/mol Because more mols of AgNO3 can be produced than are actually present, AgNO3 is the limiting reagent. n = 0.118 mol n = 0.157mol ( ) N G 2 1 (0.157 mols)(2/1) = 0.317 mol

  14. Example L.R E.R 2 1 2 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s) m = 20.0 g m = 10.0 g m = 12.7 g m = ? M = 169.88 g/mol M = 63.55 g/mol n = 0.157mol n = 0.118 mol ( ) 2 2 n = 0.118 mol Once you have determined the L.R and E.R, you can used the limiting reagent amount to calculate how much MAX Silver(Ag) will be produced.

  15. Example L.R E.R 2 1 2 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s) m = 20.0 g m = 10.0 g m = 12.7 g M = 169.88 g/mol M = 63.55 g/mol n = 0.157mol n = 0.118 mol n = 0.0981 ( ) Mols of E.R – mols from L.R = 0.157mol – 0.0589mol = 0.0981 1 2 nCu(s)Left Over= n = 0.0589 mol You can also use the amount of the L.R to determine how much of the E.R will be used up….subtract these two amounts and you can calculate how much is left over.

  16. Example L.R E.R 2 1 2 1 Cu(NO3)2 + Cu + AgNO3  Ag (s) (aq) (aq) (s) m = 20.0 g m = 10.0 g m = 12.7 g M = 169.88 g/mol M = 63.55 g/mol n = 0.157mol n = 0.118 mol n = 0.0981 Mass Cu(s) = nM = (0.0981 mol)(63.55 g/mol) = 6.23 g You can calculate the MASS of Cu left over by converting the mols into grams by multiplying by the molar mass.

  17. Example 2 L.R E.R 1 3 1 3 FeCl3(aq) + NaOH(aq)  Fe(OH)3(aq) + NaCl(aq) m = 21.5 g m = 26.8 g n = m M 0.042 molNaOH left over. n = 0.538 mol n = 0.165 mol 0.538mol – 0.496 mol = mols of NaOH left over. ( ) 3 1 n = 0.496 mol

  18. Example 2 L.R E.R 1 3 1 3 FeCl3(aq) + NaOH(aq)  Fe(OH)3(aq) + NaCl(aq) m = 21.5 g m = 26.8 g n = 0.538 mol n = 0.165 mol 0.042 molNaOH left over. n = m M m = nM m = (0.042 mol)(40.00 g/mol) = 1.7 g of NaOH left over.

  19. Example 2 L.R 1.7 g of NaOH left over. E.R 1 3 1 3 FeCl3(aq) + NaOH(aq)  Fe(OH)3(aq) + NaCl(aq) m = 17.7 g m = ? m = 21.5 g m = ? m = 26.8 g n = 0.538 mol m = nM ( ) 1 1 n = 0.165 mol n = 0.165 mol ****To determine the mass of each product you use the L.R and do stoichiometry.****

  20. Example 2 L.R 1.7 g of NaOH left over. E.R DONE 1 3 1 3 FeCl3(aq) + NaOH(aq)  Fe(OH)3(aq) + NaCl(aq) m = 29.0 g m = 21.5 g m = ? m = 26.8 g m = 17.7 g n = 0.538 mol m = nM 3 1 ( ) n = 0.495 mol n = 0.165 mol ****To determine the mass of each product you use the L.R and do stoichiometry.****

  21. Titration is an experiment where you KNOW the concentrations AND volume of one solution (mols) and you use it determine the concentration of the other solution. You set up the calculations exactly the same as a regular stoichiometry. Titrant: Is the solution you put in the burette (long glass thing). Sample: Is the solution in the beaker.

  22. In titration you know enough information about either the titrant or the sample to determine how many moles are present using n = CV. During any titration as you are adding drops from the burette (mols), there will eventually be a moment when the mols of the two solutions are equal, called the “equivalence point”. But we cant see mols, so we need some way to determine when we have equal mols so we can stop and do calculations. Different solutions each have different prosperities, like pH. So we put a pH indicator in the sample so that it will change colour when we have equal mols, telling us to “end” and the titration. This colour change is called the “End Point”.

  23. Lets Try One….It Will Make More Sense After You Try One For Real.

  24. Step 1: Balanced Chemical Equation. 1 1 2 2 H2CO3(aq)  NaCl(aq) + Na2CO3(aq) + HCl(aq)

  25. Step 2: Write down what you know. 1 1 2 2 H2CO3(aq)  NaCl(aq) + Na2CO3(aq) + HCl(aq) n = 0.0150 mol m = 1.59 g C = ? V = 0.100 L n = m M n = 1.59 g 105.99 g/mol n = 0.01500142 mol

  26. Step 2: Write down what you know. 1 1 2 2 H2CO3(aq)  NaCl(aq) + Na2CO3(aq) + HCl(aq) C = 0.150 mol/L n = 0.0150 mol C = ? V = 0.100 L C = n V = 0.0150 mol 0.100L C = 0.150 mol/L

  27. Step 2: Write down what you know. 1 1 2 2 H2CO3(aq)  NaCl(aq) + Na2CO3(aq) + HCl(aq) C = 0.150 mol/L C = ? V = 0.010 L

  28. The information above was collected after completing 4 trials of titrations. You do titrations multiple times to make sure you didn’t mix too much together and add more than you needed to. You want to take the THREE closes trials and average them….scratch the trial that is way off. This is the volume of HCl needed to get equal mols. = 12.8 ml 12.7 + 12.8 + 12.8 3 trials

  29. Step 2: Write down what you know. 1 1 2 2 H2CO3(aq)  NaCl(aq) + Na2CO3(aq) + HCl(aq) C = 0.150 mol/L C = ? V = 0.010 L V = 0.0128 L From now on its just solution stoichiometry. n = CV ( ) 2 1 n = 0.003 mol n = 0.0015 mol

  30. DONE Step 2: Write down what you know. 1 1 2 2 H2CO3(aq)  NaCl(aq) + Na2CO3(aq) + HCl(aq) C = 0.150 mol/L C = 0.234 mol/L C = ? V = 0.010 L V = 0.0128 L n = 0.003 mol C = n V C = 0.234 mol/L C = 0.003 mol 0.0128 L

  31. During a titration you are either adding: An acid to a base. A base to an acid. 12------------------------------------- If you added acid titrate to a base sample you would start at a high pH and it would gradually decrease as you added more acid, eventually leveling off when there’s lots of acid. pH 2-------------------------------------- Volume of titrate added (mL) So if you recorded the pH of the mixture of the two solution as you mixed them you would get a titration curve.

  32. Pure Base Equivalence Point 7------------------------------------ Pure Acid

  33. Equivalence Point: Is the point during a titration when the amount of acid (mols) is equal to the amount of base (mols). End Point: Is the point during a titration a COLOUR change is observed due to the presence of an indicator added to the solution. Titrant: The solution in a titration that is added, AKA the solution in the big glass tube (burette). Sample:

  34. Sodium Hydroxide titrated with Hydrochloric acid. 1 1 1 1 NaOH(aq) + HCl(aq)  NaCl(aq) + HOH(l) Na+(aq) + OH-(aq) H+(aq) + Cl-(aq)  Na+(aq) + Cl-(aq) HOHl) Total Ionic Equation OH-(aq) +  H+(aq) HOHl) ***This net ionic equation only has 1 H+(aq) and 1 OH-(aq)*** Net Ionic Equation

  35. 1. So before you begin there is just HCl, so you start the titration curve low. 2. As you start adding NaOH the pH will start to rise as the HCl gets canceled out. 3. When all the HCl has been cancelled out, the next drop of NaOH that is added causes the pH to jump. (mols are equal) pH 2------------------------------------------ Volume of NaOH added (mL)

  36. 4. At the end of the tiration when there is only base left the pH will level off at the pure base level. 12------------------------------------- 7------------------------------------ pH Equivalence Point 2------------------------------------------ Volume of NaOH added (mL)

  37. Sodium Carbonate titrated with Hydrochloric acid. 2 1 1 2 Na2CO3(aq) + HCl(aq)  NaCl(aq) + H2CO3(aq) 2 Na+(aq) + CO32-(aq) 2H+(aq) + 2Cl-(aq)  2Na+(aq) + 2Cl-(aq) H2CO3(aq) Total Ionic Equation CO32-(aq) +  2H+(aq) H2CO3(aq) ***This net ionic equation has 2H+(aq) and 1 CO32-(aq)*** Net Ionic Equation ***Because there are MORE THAN 1 H+(aq) there will be MORE THAN 1 “BUMP”.***

  38. 1. So before you begin there is just Na2CO3(aq), so you start the titration curve High. 2. As you start adding HCl the pH will start to rise as the Na2CO3(aq) gets canceled out. 11------------------------------------- 3. When all the Na2CO3 has been cancelled out, the next drop of HCL that is added causes the pH to jump. (mols are equal) pH 2------------------------------------------ Volume of HCl added (mL)

  39. 4. BECAUSE there are 2 H+, there will be 3 levels instead of two. 4. Once all the intermediate NaCO3- get cancelled out and the pH drops again. 11------------------------------------- pH 5. Once all the intermediate Na2CO3-AND NaCO3- get cancelled out and only HCl remains, the pH will level off at pure acid. 4------------------------------------ 2------------------------------------------ Equivalence Point Volume of HCl added (mL)

  40. End Points and Indicators ***When choosing an indicator for a titration choose an indicator that will change colour as close to the equivalence pH as possible.*** Strong Acid With Strong Base 7 Equivalence Point Strong Acid With Weak Base 5 Equivalence Point Weak Acid With Strong Base 10 Equivalence Point

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