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Chapter 10 Acids and Bases

Chapter 10 Acids and Bases. 10.5 Reactions of Acids and Bases. Acids and Metals. Acids react with metals such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn to produce hydrogen gas and the salt of the metal Equations: 2K( s ) + 2HCl( aq ) 2KCl( aq ) + H 2 ( g )

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Chapter 10 Acids and Bases

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  1. Chapter 10 Acids and Bases 10.5 Reactions of Acids and Bases

  2. Acids and Metals Acids react with metals • such as K, Na, Ca, Mg, Al, Zn, Fe, and Sn • to produce hydrogen gas and the salt of the metal Equations: 2K(s) + 2HCl(aq) 2KCl(aq) + H2(g) Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) Net ionic equations: 2K(s) + 2H+(aq) 2K+(aq) + H2(g) Zn(s) + 2H+(aq) Zn2+ (aq) + H2(g)

  3. Acids and Carbonates Acids react with carbonates and hydrogen carbonates to produce carbon dioxide gas, a salt, and water. 2HCl(aq) + CaCO3(s) CO2(g) + CaCl2(aq) + H2O(l) HCl(aq) + NaHCO3(s) CO2(g) + NaCl (aq) + H2O(l)

  4. Learning Check Write the products of the following reactions of acids as the complete equation and net ionic equation: A. Zn(s) + 2 HCl (aq) ? B. MgCO3 (s) + 2HCl(aq) ?

  5. Solution Write the products of the following reactions of acidsas the complete equation and net ionic equation: A. Zn(s) + 2HCl(aq) ZnCl2(aq)+ H2(g) Zn(s) + 2H+(aq)Zn2+(aq)+ H2(g) B. MgCO3(s) + 2HCl(aq) MgCl2(aq) + CO2(g) + H2O(l) MgCO3(s) + 2H+(aq)Mg2+(aq)+ H2(g)

  6. Neutralization Reactions Neutralization is the reaction of • an acid such as HCl and a base such as NaOH HCl(aq) + H2O(l) H3O+ (aq) + Cl−(aq) NaOH(aq) Na+ (aq) + OH−(aq) • the H3O+ from the acid and the OH− from the base to form water H3O+(aq) + OH−(aq) 2H2O(l)

  7. Neutralization Equations • In the equations for neutralization, an acid and a base produce a salt and water. acid base salt water HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) H+(aq) + OH−(aq) H2O 2HCl(aq) + Ca(OH)2(aq) CaCl2(aq) + 2H2O(l) H+(aq) + OH−(aq) H2O(l)

  8. Guide to Balancing an Equation for Neutralization

  9. Balancing Neutralization Reactions Write the balanced equation for the neutralization of magnesium hydroxide and nitric acid. STEP 1Write the base and acid formulas: Mg(OH)2(aq) + HNO3(aq) STEP 2Balance OH– and H+: Mg(OH)2(aq) + 2HNO3(aq) STEP 3Balance with H2O: Mg(OH)2(aq) + 2HNO3(aq) salt + 2H2O(l) STEP 4Write the salt from remaining ions: Mg(OH)2(aq) + 2HNO3(aq) Mg(NO3)2(aq) + 2H2O(l)

  10. Learning Check Select the correct group of coefficients for the following neutralization equations. A. HCl (aq) + Al(OH)3(aq) AlCl3(aq) + H2O(l) 1) 1, 3, 3, 1 2) 3, 1, 1, 1 3) 3, 1, 1, 3 B. Ba(OH)2(aq) + H3PO4(aq) Ba3(PO4)2(s)+ H2O(l) 1) 3, 2, 2, 2 2) 3, 2, 1, 6 3) 2, 3, 1, 6

  11. Solution A. 3) 3, 1, 1, 3 3HCl(aq) + Al(OH)3(aq) AlCl3(aq) + 3H2O(l) B. 2) 3, 2, 1, 6 3Ba(OH)2 (aq) + 2H3PO4(aq) Ba3(PO4)2(s) + 6H2O(l)

  12. Antacids Antacids • are used to neutralize stomach acid (HCl)

  13. Learning Check Write the neutralization reactions for stomach acid HCl and Mylanta.

  14. Solution Write the neutralization reactions for stomach acid HCl and Mylanta. STEP 1Mylanta: Al(OH)3 and Mg(OH)2 Write the base and acid formulas for each: Mg(OH)2(aq) + HCl(aq) ? Al(OH)3(aq) + HCl(aq) ? STEP 2 Balance OH- and H+ in each: Mg(OH)2(aq) + 2HCl(aq) ? Al(OH)3(aq) + 3HCl(aq) ?

  15. Solution (continued) STEP 3Balance each with H2O: Mg(OH)2(aq) + 2HCl(aq)salt + 2H2O(l) Al(OH)3(aq) + 3HCl(aq) salt + 3H2O(l) STEP 4Write the salt from remaining ions for each: Mg(OH)2(aq) + 2HCl(aq) MgCl2(aq) + 2H2O(l) Al(OH)3(aq) + 3HCl(aq) AlCl3(aq) + 3H2O(l)

  16. Acid–Base Titration Titration • is a laboratory procedure used to determine the molarity of an acid • uses a base such as NaOH to neutralize a measured volume of an acid Base (NaOH) Acid solution

  17. Indicator An indicator • is added to the acid in the flask • causes the solution to change color when the acid is neutralized

  18. End Point of Titration At the end point, • the indicator has a permanent color • the volume of the base used to reach the end point is measured • the molarity of the acid is calculated using the neutralization equation for the reaction

  19. Guide to Calculating Molarity

  20. Calculating Molarity from a Titration with a Base What is the molarity of an HCl solution if 18.5 mL of a 0.225 M NaOH are required to neutralize 10.0 mL HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 1 Given:18.5 mL (0.0185 L) of 0.225 M NaOH 10.0 mL of NaOH Need: M of HCl STEP 2 Plan: L of NaOH moles of NaOH moles of HCl M of HCl

  21. Calculating Molarity from a Titration with a Base (continued) STEP 3 State equalities and conversion factors: 1 L of NaOH = 0.225 mole of NaOH 1 L of NaOH and 0.225 mole NaOH 0.225 mole NaOH 1 L of NaOH 1 mole of NaOH = 1 mole of HCl 1 mole of NaOH and 1 mole HCl 1 mole HCl 1 mole of NaOH STEP 4 Set up the problem to calculate moles of HCl: 0.0185 L NaOH x 0.225 mole NaOH x 1 mole HCl 1 L NaOH 1 mole NaOH = 0.00416 mole of HCl

  22. Calculating Molarity from a Titration with a Base (continued) STEP 4 (continued) Calculate the volume in liters of HCl: 10.0 mL HCl = 0.0100 L HCl Calculate the molarity of HCl: 0.00416 mole HCl = 0.416 M HCl 0.0100 L HCl

  23. Learning Check Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) 1) 12.5 mL 2) 50.0 mL 3) 200 mL

  24. Solution Calculate the mL of 2.00 M H2SO4 required to neutralize 50.0 mL of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) 1) 12.5 mL STEP 1 Given:50.0 mL (0.0500 L) of 1.00 M KOH 2.0 M H2SO4 Need: M of H2SO4 STEP 2 Plan: L of KOH moles of KOH moles of H2SO4 mL of H2SO4

  25. Solution (continued) STEP 3 State equalities and conversion factors: 1 L of KOH = 1.00 mole of KOH 1 L of KOH and 1.00 mole KOH 1.00 mole KOH 1 L of KOH 2 moles of KOH = 1 mole of H2SO4 2 moles KOH and 1 mole H2SO4 1 mole H2SO4 2 moles KOH 1 L of H2SO4 = 1000 mL of H2SO4 1 L H2SO4 and 1000 mL H2SO4 1000 mL H2SO4 1 L H2SO4

  26. Solution (continued) 1 L of H2SO4 = 2.00 moles of H2SO4 1 L H2SO4 and 2.00 moles H2SO4 2.00 moles H2SO4 1 L H2SO4 STEP 4 Set up the problem to calculate the milliliters of 2.00 M H2SO4: 0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x 1 L 2 moles KOH 1 L x 1000 mL = 12.5 mL 2.00 moles H2SO4 1 L

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