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chapter11 - 11 11 Reactions in Aqueous Solutions II...

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1 Reactions in Aqueous olutions II: 11 Solutions II: Calculations 1
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hapter Goals Chapter Goals Aqueous Acid-Base Reactions 1. Calculations Involving Molarity 2. Titrations 3. Calculations for Acid-Base Titrations Oxidation Oxidation-Reduction Reactions Reduction Reactions . alancing Redox Equations 4. Balancing Redox Equations 5. Adding in H + , OH - , or H 2 O to Balance xygen or Hydrogen Oxygen or Hydrogen 6. Calculations for Redox Titrations 2
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alculations Involving Molarity Calculations Involving Molarity Example 11-1: If 100.0 mL of 1.00 M NaOH and 100.0 mL of 0.500 M H 2 SO 4 solutions are mixed, what will the concentration of the resulting solution be? hat is the balanced reaction? What is the balanced reaction? It is very important that we always use a balanced hemical reaction when doing stoichiometric chemical reaction when doing stoichiometric calculations. 3
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alculations Involving Molarity Calculations Involving Molarity eaction O H 2 SO Na SO H + NaOH 2 2 4 2 4 2 + Reaction Ratio: 2 mmol 1 mmol 1 mmol 2 mmol Before Reaction: 100 0 mmol mmol mmol mmol 50 mmol 0 mmol 0 mmol fter After Reaction: 0 mmol 0 mmol 50 mmol 100 mol 4 mmol
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alculations Involving Molarity Calculations Involving Molarity • What is the total volume of solution? 100.0 mL + 100.0 mL = 200.0 mL hat is the sodium sulfate amount in • What is the sodium sulfate amount, in mmol? 50.0 mmol hat is the molarity of the solution? What is the molarity of the solution? M = 50 mmol/200 mL = 0.250 M Na 2 SO 4 5
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alculations Involving Molarity Calculations Involving Molarity Example 11-2: If 130.0 mL of 1.00 M KOH and 100.0 mL of 0.500 M H 2 SO 4 solutions are mixed, what will be the concentration of KOH and K 2 SO 4 in the resulting solution? hat is the balanced reaction? What is the balanced reaction? 6
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alculations Involving Molarity Calculations Involving Molarity 2 KOH + H SO K SO + 2 H O 24 2 eaction Reaction Ratio: 2 mmol 1 mmol 1 mmol 2 mmol Before Reaction: 130 50 mmol 0 mmol 0 mmol mmol fter After Reaction: 30 mol 0 mmol 50 mmol 100 mol 7 mmol mmol
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alculations Involving Molarity Calculations Involving Molarity • What is the total volume of solution? 30 0 mL + 100 0 mL = 230 0 mL 130.0 mL + 100.0 mL 230.0 mL • What are the potassium hydroxide and potassium sulfate amounts? 0.0 mmol & 50.0 mmol 30 0 o & 50 0 o • What is the molarity of the solution? M = 30.0 mmol/230.0 mL = 0.130 M KOH = 50.0 mmol/230.0 mL = 0.217 M K O 8 M 50.0 mmol/230.0 mL 0.217 M K 2 SO 4
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alculations Involving Molarity Calculations Involving Molarity Example 11-3: What volume of 0.750 M NaOH solution would be required to completely neutralize 100 mL of 0.250 M H 3 PO 4 ? O a O aOH PO H mol 0.250 O 100 = aOH O H 3 + PO Na PO H + NaOH 3 4 3 2 4 3 4 3 aOH aOH ol PO H L 1 PO H L 0.100 NaOH L ? 4 3 4 3 × × NaOH L 0.100 NaOH mol 0.750 NaOH L 1 PO H mol 1 NaOH mol 3 = × 9 4 3
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itrations Titrations Acid-base Titration Terminology ± Titration A method of determining the concentration of one solution by reacting it with a solution of known concentration.
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chapter11 - 11 11 Reactions in Aqueous Solutions II...

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