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Chapter 14 Acids and Bases

Chapter 14 Acids and Bases. 14.8 Acid-Base Titration. Titration http://www.youtube.com/watch?v=g8jdCWC10vQ&list=PL13E6D5AEF3861CE0&index=8&feature=plpp_video. Acid-Base Titration. Base (NaOH). Titration is a laboratory procedure used to determine the molarity of an acid

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Chapter 14 Acids and Bases

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  1. Chapter 14 Acids and Bases 14.8 Acid-Base Titration

  2. Titration • http://www.youtube.com/watch?v=g8jdCWC10vQ&list=PL13E6D5AEF3861CE0&index=8&feature=plpp_video

  3. Acid-Base Titration Base (NaOH) 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 Acid solution

  4. Indicator An indicator • is added to the acid in the flask • changes the color of the solution when the acid is neutralized

  5. 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

  6. Calculations for an Acid-Base Titration

  7. Example of Calculating Molarity What is the molarity of 10.0 mL of HCl solution if 18.5 mL of 0.225 M NaOH are required to neutralize the HCl? HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) STEP 1State the given and needed quantities and concentration. Given 18.5 mL of a 0.225 M NaOH solution 18.5 mL of NaOH = 0.0185 L of NaOH 10.0 mL of HCl = 0.0100 L of HCl Need molarity of HCl solution

  8. Calculating Molarity (continued) STEP 2Write a plan to calculate molarity. L of NaOH moles of NaOH moles of HCl molarity of HCl STEP 3 State equalities and conversion factors including concentrations. 1 L of NaOH = 0.225 mol of NaOH 1 L NaOH and 0.225 mol NaOH 0.225 mol NaOH 1 L NaOH 1 mol of NaOH = 1 mol of HCl 1 mol HCl and 1 mol NaOH 1 mol NaOH 1 mol HCl

  9. Calculating Molarity (continued) STEP 4 Set up problem to calculate needed quantity. 0.0185 L NaOH x 0.225 mol NaOH x 1 mol HCl 1 L NaOH 1 mol NaOH = 0.00416 mol of HCl Molarity of HCl = 0.00416 mol HCl = 0.416 M HCl 0.0100 L HCl

  10. Learning Check Calculate the milliliters of 2.00 M H2SO4 needed to completely neutralize 0.0500 L of 1.00 M KOH. H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l) 1) 12.5 mL 2) 50.0 mL 3) 200. mL

  11. Solution STEP 1State the given and needed quantities and concentration. Given 0.0500 L of a 1.00 M KOH solution 1 L of H2SO4 = 2.00 mol of H2SO4 1000 mL of H2SO4 = 1 L of H2SO4 Need milliliters of H2SO4 solution STEP 2Write a plan to calculate milliters. L of KOH moles of KOH moles of H2SO4 liters of H2SO4 mL of H2SO4

  12. Solution (continued) STEP 3 State equalities and conversion factors including concentrations. 1 L of KOH = 1.00 mol of KOH 1 L KOH and 1.00 mol KOH 1.00 mol KOH 1 L KOH 2 mol of KOH = 1 mol of H2SO4 1 mol H2SO4 and 2 mol KOH 2 mol KOH 1 mol H2SO4 1 L of KOH = 1000 mL of KOH 1 L KOH and 1000 mL KOH 1000 mL KOH 1 L KOH

  13. Solution (continued) STEP 3 (continued) 1 L of H2SO4 = 2.00 mol of H2SO4 1 L H2SO4 and 2.00 mol H2SO4 2.00 mol H2SO4 1 L H2SO4 STEP 4 Set up problem to calculate needed quantity. 0.0500 L KOH x 1.00 mol KOH x 1 mol H2SO4 1 L KOH 2 mol KOH x 1 L H2SO4 x 1000 mL H2SO4 = 12.5 mL of H2SO4 2.00 mol H2SO4 1 L H2SO4 (1)

  14. Learning Check A 25.0 mL sample of phosphoric acid is neutralized by 42.6 mL of 1.45 M NaOH. What is the molarity of the phosphoric acid solution? 3NaOH(aq) + H3PO4 (aq) Na3PO4(aq) + 3H2O(l) 1) 0.62 M 2) 0.841 M 3) 0.185 M

  15. Solution STEP 1State the given and needed quantities and concentration. Given 42.6 mL of a 1.45 M NaOH solution 25.0 mL of a H3PO4 solution Need molarity of H3PO4 solution STEP 2Write a plan to calculate molarity. mL of NaOH L of NaOH moles of NaOH moles of H3PO4 liters of H3PO4 molarity of H3PO4

  16. Solution (continued) STEP 3 State equalities and conversion factors including concentrations. 1 L of NaOH = 1.45 mol of NaOH 1 L NaOH and 1.45 mol NaOH 1.45 mol NaOH 1 L NaOH 3 mol of NaOH = 1 mol of H3PO4 1 mol H3PO4 and 3 mol NaOH 3 mol NaOH 1 mol H3PO4 1 L of NaOH = 1000 mL of NaOH 1 L NaOH and 1000 mL NaOH 1000 mL NaOH 1 L NaOH

  17. Solution (continued) STEP 3 (continued) 1 L of H3PO4 = 1000 mL of H3PO4 1 L H3PO4 and 1000 mL H3PO4 1000 mL H3PO4 1 L H3PO4 STEP 4 Set up problem to calculate needed quantity. 0.0426 L NaOH x 1.45 mol NaOH x 1 mol H3PO4 1 L NaOH3 mol NaOH = 0.0206 mol of H3PO4

  18. 25.0 mL H3PO4 x1L H3PO4 = 0.0250 L H3PO4 1000 mL H3PO4 0.0206 mol H3PO4 = 0.841 M (2) 0.0250 L H3PO4 Solution (continued) 18

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