1 / 65

Reactions in Aqueous Solutions: Calculations

Reactions in Aqueous Solutions: Calculations . Concentration….is what?. Imagine you put some orange squash in a 500ml glass and fill it with water. Now put the same quantity in a 1000ml flask What would be the difference??. How do we measure conc?.

carr
Download Presentation

Reactions in Aqueous Solutions: Calculations

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Reactions in Aqueous Solutions: Calculations

  2. Concentration….is what? • Imagine you put some orange squash in a 500ml glass and fill it with water. • Now put the same quantity in a 1000ml flask • What would be the difference??

  3. How do we measure conc? • We work out how many moles there are in every litre! • Units of volume?? • dm3 which is the same as a litre (1000cm3) • UNITS = mol dm3 (or M for short)

  4. How does it work • Easy! If you have 2 moles of Bromine and dissolve it in 1 dm3 of water the conc. would be? • 2mol dm3 (Chemists often say that this is a 2 Molar solution) • What about if we dissolved it in 2 dm3?? • 1mol dm3

  5. n c v Time for an equation to help! • n=cV VOLUME in dm3 dm3= cm3 / 1000 Pg 14 Q 1 Q3 every other

  6. Acid-Base Titration Terminology • titration • primary standard • standard solution • standardization • indicator • equivalence point • end point

  7. Titrations • The process in which a solution of one reactant, the titrant, is carefully added to a solution of another reactant, and the volume of titrant required for complete reaction is measured. • It lets you work out concentration (amongst other things!) • E.g. the amount of acid in hair shampoo.

  8. The Buret - Volumetric glassware used for titrations. The buret allows you measure the exact amount of the titrant to the solution you are testing. An indicator in the solution being tested tells when the endpoint has been reached. A pH meter will allow for the equivalence point to be determined. Titrations:Experimental Method

  9. Titrations • A standard solution is a solution which the concentration is accurately known.

  10. Add solution from the buret. Reagent (base) reacts with compound (acid) in solution in the flask. Indicator shows when exact stoichiometric reaction has occurred. Net ionic equation: H+ + OH-H2O At equivalence point moles H+ = moles OH- Titration

  11. Titrations • Equivalence point is the point at which chemically equivalent amounts of reactants have reacted. • End point is the point at which an indicator changes color and the titration is stopped.

  12. Titrations • Equivalence point is the point at which chemically equivalent amounts of reactants have reacted. • End point is the point at which an indicator changes color and the titration is stopped.

  13. What use is it to us?? • We know the concentration and volume of our measured sample (the acid) • Therefore we can work out the number of … • Moles!! • Assuming a 1:1 ratio there must be as many moles of the acid as of the alkali, so now we know how many moles of alkali we have too! • We know the volume of the alkali from the burette as well so we can calculate its concentration

  14. A worked example • Example 12.1: Given the equation NaOH(aq) + HCl(aq) ==> NaCl(aq) + H2O(l) • 25 cm3 of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.2M hydrochloric acid. Using a suitable indicator it was found that 15 cm3 of acid was required to neutralise the alkali. Calculate the molarity (concc) of the sodium hydroxide.

  15. moles HCl = (15/1000) x 0.2 = 0.003 mol • moles HCl = moles NaOH (1 : 1 in equation) • so there is 0.003 mol NaOH in 25 cm3 • scaling up to 1000 cm3 (1 dm3), there are ... • 0.003 x (1000/25) = 0.12 mol NaOH in 1 dm3 • molarity of NaOH is 0.12M or mol dm-3  • since mass = moles x formula mass, and Mr(NaOH) = 23 + 16 + 1 = 40 • concentration in g/dm3 is 0.12 x 40 = 4.41g/dm3 

  16. A solution of sodium hydroxide contained 0.25 mol dm-3. Using phenolphthalein indicator, titration of 25.0 cm3 of this solution required 22.5 cm3 of a hydrochloric acid solution for complete neutralisation.

  17. a) write the equation for the titration reaction. • (b) what apparatus would you use to measure out (i) the sodium hydroxide solution? (ii) the hydrochloric acid solution? • (c) what would you rinse the apparatus out with before doing the titration ? • (d) what is the indicator colour change at the end-point? • (e) calculate the moles of sodium hydroxide neutralised. • (f) calculate the moles of hydrochloric acid neutralised. • (g) calculate the concentration of the hydrochloric acid in mol/dm3 (molarity).

  18. A solution of pure barium hydroxide contained 2.74 g in exactly 100 cm3 of water. Using phenolphthalein indicator, titration of 20.0 cm3 of this solution required 18.7 cm3 of a hydrochloric acid solution for complete neutralisation.

  19. (a) write the equation for the titration reaction. • (b) calculate the molarity of the barium hydroxide solution. • (c) calculate the moles of barium hydroxide neutralised. • (d) calculate the moles of hydrochloric acid neutralised. • (e) calculate the molarity of the hydrochloric acid.

  20. 4.90g of pure sulphuric acid was dissolved in water, the resulting total volume was 200 cm3. 20.7 cm3 of this solution was found on titration, to completely neutralise 10.0 cm3 of a sodium hydroxide solution.

  21. (a) write the equation for the titration reaction. • (b) calculate the molarity of the sulphuric acid solution. • (c) calculate the moles of sulphuric acid neutralised. • (d) calculate the moles of sodium hydroxide neutralised. • (e) calculate the concentration of the sodium hydroxide in mol dm-3 (molarity).

  22. Titration Stoichiometry • What is the molarity of a HCl solution if 36.7 mL of the HCl is required to react with 43.2 mL of 0.236 M NaOH? HCl + NaOH  NaCl + H2O

  23. Titration Stoichiometry HCl + NaOH  NaCl + H2O 36.7mL 43.2mL 0.236M Mol HCl = 0.0432 L NaOH = 0.0102 mol HCL

  24. Titration Stoichiometry • A 43.2 mL sample of 0.236 M NaOH reacts completely with 36.7 mL of a sulfuric acid solution. • What is the molarity of the H2SO4 solution? 2NaOH + H2SO4 Na2SO4 + H2O

  25. Titration Stoichiometry 2NaOH + H2SO4 Na2SO4 + H2O 43.2 mL 0.236 M 36.7 mL ? M mol H2SO4 = 0.0432 L NaOH = 0.00510 mol H2SO4

  26. Calculations Involving Molarity • If 100 mL of 1.00 M NaOH and 100 mL of 0.500 M H2SO4 solutions are mixed, what will the concentration of the resulting solution be? • What is the balanced reaction?

  27. Calculations Involving Molarity before rxn: 100mmol 50mmol 0mmol 0mmol after rxn: 0mmol 0mmol 50mmol 100mmol

  28. Calculations Involving Molarity • What is the total volume of solution? • 100 mL + 100 mL = 200 mL • What is amount of sodium sulfate? • 50.0 mmol • What is the molarity of the solution? • M = 50 mmol/200 mL = 0.250 M Na2SO4

  29. Titration • Add solution from the buret. • Reagent (base) reacts with compound (acid) in solution in the flask. • Indicator shows when exact stoichiometric reaction has occurred. • Net ionic equation: • H+ + OH-H2O • At equivalence point moles H+ = moles OH-

  30. Standardization of Acid–Base Solutions • Standardization is the process by which one determines the concentration of a solution by measuring the volume of the solution required to react with a known amount of a primary standard. • The standardized solution is known as a secondary standard.

  31. Properties of an Ideal Primary Standard • It must not react with or absorb the components of the atmosphere, such as water vapor, oxygen, and carbon dioxide. • It must react according to one invariable reaction. • It must have high percentage purity.

  32. Properties of an Ideal Primary Standard • It should have a high formula weight to minimize the effect of weighing error. • It must be soluble in the solvent of interest, usually water. • It should be nontoxic.

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

  34. Examples • How many grams of Ba(OH)2 are required to prepare 2.50 L of 0.06000M? g = 2.5 L = 25.7 g

  35. 2HCl + Ca(OH)2 Ž 2H2O + CaCl2 Examples • How many mL of 0.00300M HCl are needed to neutralize 30.0 mL of 0.00100M Ca(OH)2? mL HCl = 0.0300 L Ca(OH)2 = 20.0 mL HCl

  36. M = 0.00343 mol Examples • What is the M of H2SO4 if 40.0 mL neutralized 0.364 g Na2CO3? H2SO4 + Na2CO3Ž Na2SO4 + CO2 + H2O mol H2SO4 = 0.364 g Na2CO3 = 0.00343 mol = 0.0858 M 0.0400 L

  37. Mole Method & Molarity • potassium hydrogen phthalate • KHP - molar mass = 204.2 g/mol

  38. M = 0.001783 mol Examples • What is the M of NaOH if 20.00 mL neutralize 0.3641g KHP? NaOH + KHP Ž H2O + NaKP mol NaOH = 0.3641 g KHP = 0.001783 mol = 0.08951 M 0.02000 L

  39. Examples • What is the percent purity of oxalic acid if 39.82 mL of 0.08915 M NaOH neutralized 0.1743 g of an impure oxalic acid sample? g Hox = 0.03982 L NaOH = 91.68% Hox % purity =

  40. Examples • How many mL of 0.0200 M KMnO4 are needed to react with 40.0 mL of 0.100 M Fe2+? MnO4- + 8H+ + 5Fe2+Ž 5Fe3+ + Mn2+ + 4H2O mL MnO4- = 0.0400L Fe2+ =40.0 mL MnO4-

  41. REDOX Reactions OilRig • Oxidations - Loss of electrons • Reduction - Gain of electrons Cu Ž Cu+2 + 2e Cu+2 + 2e Ž Cu

  42. Balancing Redox Reactions • Divide the equation into two incomplete half–reactions. • First, balance the elements other than H and O. • Next, balance the O atoms by adding H2O. • Then, balance the H atoms by adding H+.

  43. Balancing Redox Reactions • Finally, balance the charge by adding e. • Multiply each half–reaction by an integer so that the number of electrons lost in one half–reaction equals the number gained in the other. • Add the two half–reactions. • Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.

  44. +2 -1 0 +1 -1 0 oxidized reduced 2 2 2 2Ag + CuCl2 Ž 2AgCl + Cu Example CuCl2 + Ag Ž AgCl+ Cu 2x) Ag Ž Ag+ + e 2e + Cu+2Ž Cu

  45. 0 0 +3 -1 6 / 2 2 2x) 6 / 6 / 3 3x) 2Fe + 3Cl2 Ž 2FeCl3 Example Fe + Cl2Ž FeCl3 + 3e Fe Ž Fe+3 2e + Cl2 Ž Cl- Cl2 Ž 2 Cl-

  46. +1 -1 0 0 +3 -1 oxidized reduced 6 / 2 2 6 / 6 / 3 2Al + 6HClŽ 2AlCl3 + 3H2 Example HCl + Al Ž H2 + AlCl3 2x) Al Ž Al+3 + 3e 3x) 2e + 2H+Ž H2

  47. +7 -2 -1 +2 0 oxidized reduced 10 / 10 / 5 5x) 10 / 16 / 8 / 2 2 2x) 16H++ 2MnO4- + 10Br-Ž2Mn2+ + 5Br2 + 8H2O Example MnO4- + Br-ŽMn2+ + Br2 2Br-Ž Br2 + 2e 5e + 8H+ + MnO4-Ž Mn2+ + 4H2O

  48. +6 -2 -1 +3 0 oxidized reduced 6 / 6 / 3 3x) 14H++ Cr2O72- + 6I-Ž2Cr3+ + 3I2 + 7H2O Example Cr2O72- + I-Ž Cr3+ + I2 2I-Ž I2 + 2e 6e + 14H+ + Cr2O72-Ž 2Cr3+ + 7H2O

  49. Oxidation-Reduction Reactions • Tin (II) ions are oxidized to tin (IV) by bromine. Use the half reaction method to write and balance the net ionic equation. Sn2+ + Br2 Sn4+ + Br-

More Related