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Metal Salt Displacement Activity Series Explanation

Understand metal displacement reactions, molarity, solution preparation, dilution, and stoichiometry in chemistry with clear examples. Learn ion concentrations and solve related problems.

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Metal Salt Displacement Activity Series Explanation

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  1. Metal + Acid Displacement

  2. Activity Series of Metals

  3. Activity Series of Metals • metals higher in series react with compounds of those below • metals become less reactive to water top to bottom • metals become less able to displace H2 from acids top to bottom

  4. Potassium + Water

  5. Activity Series of Metals Zn(s) + CuSO4(aq) ZnSO4(aq) + Cu(s) Cu(s) + 2AgNO3(aq) Cu(NO3)2(aq) + 2Ag(s) Fe(s) + 2HCl(aq) FeCl2(aq) + H2(g) Zn(s) + 2HBr(aq) ZnBr2(aq) + H2(g)

  6. Metal + Metal Salt Displacement

  7. 130 10 0 0 Which of the following reactions does NOT happen? • Cu(s)+H2SO4(aq)CuSO4(aq)+H2(g) • 2HNO3(aq)+2K(s)2KNO3(aq)+H2(g) • FeCl2(aq)+Zn(s)ZnCl2(s)+Fe(s) • Ca(s)+2H2O(l)Ca(OH)2(aq)+H2(g) • Cu(s)+2AgNO3(aq)2Ag(s)+Cu(NO3)2(aq)

  8. Solution A homogeneous mixture of two or more substances comprising the solvent which is the majority of the mixture and one or more solutes which are the smaller fraction. Concentration – how much solute is present in a given amount of solution

  9. Cola Drinks Solvent • water Solutes • carbon dioxide (gas) • sweetener (solid) • phosphoric acid (liquid) • caramel color (solid)

  10. Molarity – a measure of concentration The number of moles of solute per liter of solution. molarity  M moles of solute M = liters of solution units  molar = mol/L = M

  11. Preparation of 1.00 L of 0.0100 M KMnO4 solution from solid

  12. 130 10 0 0 Which value do you NOT need to determine the molarity of a solution? • Mass of solute • Molar mass of solute • Volume of solvent added • Total volume of solution

  13. Solution Preparation by Dilution

  14. Dilution • Molarity (mol/L) × Volume (L) = Moles • If we take a sample (say 25.0 mL) from a solution (say 0.372 M) and add extra water (say to a total volume of 500. mL) the moles of solute are unchanged • Thus M1V1 = moles = M2V2 • 0.372 M × 25.0 mL = M2 × 500. mL • M2 = 0.0186 M

  15. 130 10 0 0 How many L of conc HNO3 (16.0 M) are needed to prepare 0.500 L of 0.250 M nitric acid? • 32.0 L • 16.0 L • 0.500 L • 0.250 L • 0.00781 L

  16. Stoichiometric Relationships

  17. Titrationsat equivalence mol H+ = mol OH-

  18. EXAMPLE: A sample of lye, sodium hydroxide, is neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH? 2 NaOH(aq) + H2SO4(aq) Na2SO4(aq) + 2 H2O (25.0 mL NaOH) #mL H2SO4 = (1 L) (1000 mL) (0.400 mol NaOH) (1 L NaOH) (1000 mL) (1 L) (1 L H2SO4) (0.200 mol H2SO4) (1 mol H2SO4) (2 mol NaOH) = 25.0 mL H2SO4

  19. Ion Concentrations • 0.100 M NaCl • NaCl(aq) Na+(aq) + Cl-(aq) • Is 0.100 M in both Na+ and Cl- • 0.100 M Al2(SO4)3 • Al2(SO4)3(aq)  2 Al3+(aq) + 3 SO42-(aq) • Is 0.200 M in Al3+ and 0.300 M in SO42-

  20. Final Ion Concentrations • Consider the NaOH + H2SO4 reaction. What are the final concentrations of Na+ and SO42-? • Stoichiometric reaction, can use either reactant to determine moles of product • 25.0 mL NaOH × 0.400 mol NaOH × 1 L 1 L 1000 mL0.0100 mol NaOH × 1 mol Na2SO4 = 0.0050 mol Na2SO4 2 mol NaOH

  21. Total volume = 25.0 mL + 25.0 mL = 50.0 mL = 0.0500 L • [Na2SO4] = 0.00500 mol = 0.100 M 0.0500 L • [Na+] = 2 × 0.100 M = 0.200 M • [SO42-] = 0.100 M

  22. 130 10 0 0 Which solution has the highest concentration of SO42-? • 0.20 M CuSO4 • 0.15 M Na2SO4 • 0.070 M Fe2(SO4)3 • 0.10 M Ce(SO4)2

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