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Chemistry Chapter 21 NeutralizationReactions
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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 + ) 1 M Al(OH) 3 = 3 N Al(OH) 3 (3 equivalent OH - )
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Molarity (M)Normality (N ) Moles of equivalent H + or OH - 3 M NaOH3 N NaOH3 equivalent OH - 2 M H 3 PO 4 6 N H 3 PO 4 6 equivalent H + 6 M H 2 SO 4 12 N H 2 SO 4 12 equivalent H + 5 M HNO 3 5 N HNO 3 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 - 6 M HC 2 H 3 O 2 6 N HC 2 H 3 O 2 6 equivalent H + 0.5 M Al(OH) 3 1.5 N Al(OH) 3 1.5 equivalent OH -
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Molarity (M)Normality (N ) Moles of equivalent H + or OH - 2 M NaOH2 N NaOH2 equivalent OH - 2 M H 3 PO 4 6 N H 3 PO 4 6 equivalent H + 6 M H 2 SO 4 12 N H 2 SO 4 12 equivalent H + 5 M HNO 3 5 N HNO 3 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 - 6 M HC 2 H 3 O 2 6 N HC 2 H 3 O 2 6 equivalent H + 0.4 M Al(OH) 3 1.2 N Al(OH) 3 1.2 equivalent OH -
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Molarity (M) = Normality (N ) / equivalents = HCl (20.0 mL) N A V A = N B V B NaOH (0.20 N)NANA (30.0 mL) (20.0 mL) = (0.20 N) NANA (30.0) (20.0) = NANA 0.30 N HCl = 0.30 M HClMAMA 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 M NaOH stock solution. What is the concentration (Molarity) of HCl in mol/L?
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Molarity (M) = Normality (N ) / equivalents = H 2 SO 4 (50.0 mL) N A V A = N B V B NaOH (0.20 N)NANA (10.0 mL) (50.0 mL) = (0.20 N) NANA (10.0) (50.0) = NANA 0.040 N H 2 SO 4 = 0.020 M H 2 SO 4 MAMA Example 2: In order to determine the concentration of an unknown H 2 SO 4 solution, 50.0 mL of the H 2 SO 4 is titrated using 10.0 mL of a 0.20 M NaOH stock solution. What is the concentration (Molarity) of H 2 SO 4 in mol/L?
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Lab 24 Titration Results for 10.0 mL of HCl = HCl (10.0 mL) N A V A = N B V B NaOH (0.09860 N)NANA (12.35 mL) (10.0 mL) = 0.30 N HCl M A = 0.30 M HCl 0.09860 M NaOH = 0.09860 N NaOH Equivalence point: Point where [H + ] = [OH - ] Also called the “End Point” = (12.30+12.40)/2 = 12.35 mL Molarity (M) = Normality (N ) / equivalents
![Lab 24 Titration Results for 10.0 mL of HCl = HCl (10.0 mL) N A V A = N B V B NaOH ( N)NANA (12.35 mL) (10.0 mL) = 0.30 N HCl M A = 0.30 M HCl M NaOH = N NaOH Equivalence point: Point where [H + ] = [OH - ] Also called the End Point = ( )/2 = mL Molarity (M) = Normality (N ) / equivalents](http://images.slideplayer.com/32/10002655/slides/slide_8.jpg "Lab 24 Titration Results for 10.0 mL of HCl = HCl (10.0 mL) N A V A = N B V B NaOH ( N)NANA (12.35 mL) (10.0 mL) = 0.30 N HCl M A = 0.30 M HCl M NaOH = N NaOH Equivalence point: Point where [H + ] = [OH - ] Also called the End Point = ( )/2 = mL Molarity (M) = Normality (N ) / equivalents")
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