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Chemistry Chapter 21

Chemistry Chapter 21. Neutralization Reactions. Concentration : Describe amount of dissolved solute in solution. Molarity (M) : Moles solute Liters solution. Normality ( N ) : Moles of equivalents Liters of Solution 1 M H 2 SO 4 = 2 N H 2 SO 4 (2 equivalent H + )

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Chemistry Chapter 21

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  1. Chemistry Chapter 21 Neutralization Reactions

  2. Concentration: Describe amount of dissolved solute in solution. • Molarity (M) : Moles solute • Liters solution. • Normality (N ) : Moles of equivalents • Liters of Solution • 1 M H2SO4 = 2 N H2SO4 (2 equivalent H+) • 1 M Al(OH)3 = 3 N Al(OH)3(3 equivalent OH-)

  3. Moles of equivalent H+ or OH- Molarity (M) Normality (N ) 3 M NaOH 3 N NaOH 3 equivalent OH- 2 M H3PO4 6 N H3PO4 6 equivalent H+ 6 M H2SO4 12 N H2SO4 12 equivalent H+ 5 M HNO3 5 N HNO3 5 equivalent H+ 2 M Ca(OH)2 4 N Ca(OH)2 4 equivalent OH- 2 M Al(OH)3 6 N Al(OH)3 6 equivalent OH- 0.5 M Al(OH)3 1.5 N Al(OH)3 1.5 equivalent OH- 6 M HC2H3O2 6 N HC2H3O2 6 equivalent H+

  4. Moles of equivalent H+ or OH- Molarity (M) Normality (N ) 2 M NaOH 2 N NaOH 2 equivalent OH- 2 M H3PO4 6 N H3PO4 6 equivalent H+ 6 M H2SO4 12 N H2SO4 12 equivalent H+ 5 M HNO3 5 N HNO3 5 equivalent H+ 2 M Ca(OH)2 4 N Ca(OH)2 4 equivalent OH- 2 M Al(OH)3 6 N Al(OH)3 6 equivalent OH- 0.4 M Al(OH)3 1.2 N Al(OH)3 1.2 equivalent OH- 6 M HC2H3O2 6 N HC2H3O2 6 equivalent H+

  5. Example 1: In order to determine the concentration of an unknown HCl solution, 20.0 mL of the HCl is titrated using 30.0 mL of a 0.20 MNaOH stock solution. What is the concentration (Molarity) of HCl in mol/L? Molarity (M) = Normality (N ) / equivalents HCl NaOH N AVA=N BVB = NA (20.0 mL) (0.20 N) (30.0 mL) (20.0 mL) (20.0 mL) (0.20 N) (30.0) = NA (20.0) = 0.30 N HCl NA = MA 0.30 M HCl

  6. Example 2: In order to determine the concentration of an unknown H2SO4 solution, 50.0 mL of the H2SO4 is titrated using 10.0 mL of a 0.20 M NaOH stock solution. What is the concentration (Molarity) of H2SO4 in mol/L? Molarity (M) = Normality (N ) / equivalents H2SO4 NaOH N AVA=N BVB = NA (50.0 mL) (0.20 N) (10.0 mL) (50.0 mL) (50.0 mL) (0.20 N) (10.0) = NA (50.0) = 0.040 N H2SO4 NA = MA 0.020 M H2SO4

  7. Lab 24 Titration Results for 10.0 mL of HCl Equivalence point: Point where [H+] = [OH-] Also called the “End Point” = (12.30+12.40)/2 = 12.35 mL 0.09860 M NaOH = 0.09860 N NaOH Molarity (M) = Normality (N ) / equivalents HCl NaOH MA= 0.30 M HCl N AVA=N BVB = NA (10.0 mL) (0.09860 N) (12.35 mL) = 0.30 N HCl (10.0 mL) (10.0 mL)

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