The Chemistry of Acids and Bases

Acids and Bases: Titration At the conclusion of our time together, you should be able to: 1. Do a titration of an acid with a base 2. Calculate the equivalence point (neutralization point) for a titration problem 3. Define normality

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ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using TITRATION. Oxalic Acid, H 2 C 2 O 4

Setup For Titrating An Acid With A Base

TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called the NEUTRALIZATION point. This is called the NEUTRALIZATION point. The equivalence point has been reached

The Formula for Titration Use: Use: M 1 * E * V 1 = M 2 * E * V 2 M = molarity of acid and base E = number of H+ or OH- per mole of acid or base M x E = Normality V = volume

Various Protic Titration Page 56:1 Standardize a solution of NaOH — i.e., accurately determine its concentration mL of M NaOH is neutralized with 25.0 mL of HCl by titration to an equivalence point. What is the concentration of the acid?

How do you do this?? Use: Use: M 1 * E * V 1 = M 2 * E * V M x 1 x 32.7 mL = M x 1 x 25.0 mL = M HCl pH= -log (0.131 x 1) = 0.883

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Page 56-2: You have 50.0 mL of M NaOH neutralized 20.0 mL of sulfuric acid. Determine the M of the acid? M x 1 x 50.0 mL = M x 2 x 20.0 mL = M H 2 SO 4 For each mole of sulfuric acid there are 2 moles of H +, therefore, pH = -log (0.313 x 2) = 0.203

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Acids and Bases: Titration Let’s see if you can: 1. Do a titration of an acid with a base 2. Calculate the equivalence point (neutralization point) for a titration problem 3. Define normality

Page 56:3 Various Protic Use: Use: M 1 * E * V 1 = M 2 * E * V M x 2 x 29.6 mL = M x 1 x 40.0 mL = M HNO 3 pOH= -log ( x 2) = 1.00

Page 58:1 Titration with Grams Use: Use: M 1 * E * V 1 = M 2 * E * V 2 Moles of base (1)=.800 M x 1 x L = M HCl grams (1 mol/MM) (E) = mol grams= (0.0800) 4.49 g KOH

Page 59:4 Titration with Grams Use: Use: M 1 * E * V 1 = M 2 * E * V 2 3 x Moles of acid= M x 1 x L = M KOH multiply both sides by MM and divide by MEV 3 x 2.73 g mol = Molar Mass 178 g/mol

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Normality is defined as 1. M * V 2. E * V 3. M * E 4. M * E * V 5. Not listed

A method used to determine the concentration of an unknown acid or base. 1. Equivalence point 2. Neutralization point 3. Titration 4. Titrant 5. Normality 6. Flask 7. Buret 8. Not listed

The titrant is usually added to a(n) 1. Equivalence point 2. Neutralization point 3. Titration 4. Titrant 5. Normality 6. Flask 7. Buret 8. Not listed

The indicator is usually added to the 1. Equivalence point 2. Neutralization point 3. Titration 4. Titrant 5. Normality 6. Flask 7. Buret 8. Not listed

Going Down!!

3 M sulfuric acid would have a normality of 1. 3 N 2. 2 N 3. 5 N 4. 6 N 5. 9 N 6. Not listed

100.0 mL of 3.0 M sulfuric acid neutralizes 50.0 mL of sodium hydroxide. What is the molarity of the base? M M M M M 6. Not listed

100.0 mL of M nitric acid neutralizes M of aluminum hydroxide. How many mL of the base did you use? mL mL mL mL mL 6. Not listed

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The Chemistry of Acids and Bases

Acids and Bases: Titration At the conclusion of our time together, you should be able to: 1. Do a titration of an acid with a base 2. Calculate the equivalence point (neutralization point) for a titration problem 3. Define normality 4. Determine the percent acid with a titration problem

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During a titration, the indicator goes in the 1. Erlenmyer flask 2. Beaker 3. Buret

During a titration, the substance of unknown concentration goes in the 1. Erlenmyer flask 2. Beaker 3. Buret

Lemonade Titration Problem MEV (base) = MEV (acid) M x 1 x 8.70 mL = M x 3 x 10.0 mL = M citric acid

Lemonade Titration Problem M = mol/L M x L = x g g/mol = g in 10 mL

Lemonade Titration Problem Using the density of citric acid in lemonade: 10.0 mL x g/mL = 9.60 g lemonade in 10.0 mL of solution g citric acid in 10.0 mL x 100 = 9.60 g lemonade in 10.0 mL 0.580%