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Neutralization and Titrations!

Neutralization and Titrations!. Neutralization Reactions. Reactions in which an acid and a base react in an aqueous solution to produce a salt and water. HCl + NaOH  NaCl + H 2 O. What types of reactions are these?. Double Displacement. MEMORIZE THIS:.

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Neutralization and Titrations!

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  1. Neutralization and Titrations!

  2. Neutralization Reactions • Reactions in which an acidand a basereact in an aqueous solution to produce a saltand water. HCl + NaOH  NaCl + H2O What types of reactions are these? Double Displacement

  3. MEMORIZE THIS: Acid + Base  Salt + Water

  4. YOUR TURN: Complete & Balance the reactions. Then, circle & name the salt for each reaction. ___H3PO4 + ___ Ca(OH)2 2 3 ___H2O + ___Ca3(PO4)2 6 Calcium phosphate ___H2SO4 + ___ KOH  2 ___H2O + ___K2SO4 2 Potassium sulfate

  5. TITRATIONS!

  6. Titrations! What is a titration? The process of adding a knownamountof solution of known concentrationto determine the concentration of another solution. Involves the addition of an acid to a base or a base to an acid.

  7. Equivalence Point When the number of moles of hydrogenions equals the number of moles of hydroxideions. HCl + NaOH  NaCl + H2O 1 mole 1 mole 1 mole 1 mole H2SO4 + 2NaOH  Na2SO4 + 2H2O 2 mole 1 mole 2 mole 1 mole 2HCl + Ca(OH)2 CaCl2 + 2H2O 1 mole 2 mole 2 mole 1 mole

  8. How many moles of HCl will it take to neutralize 6 moles of NaOH? How many moles of HCl would it take to neutralize 6 moles of Ca(OH)2? How many moles of NaOH would it take to neutralize 6 moles of H2SO4? How many moles of H3PO4 would it take to neutralize 9 moles of Ca(OH)2? How many moles of H3PO4 would it take to neutralize 6.4 moles of Ca(OH)2? 6 moles HCl 12 moles HCl 12 moles NaOH 6 moles H3PO4 2 H3PO4 + 3 Ca(OH)2 6 H2O + Ca3(PO4)2 ) = ( 6.4 mol Ca(OH)2 2 mol H3PO4 4.3 mol 3 mol Ca(OH)2 H3PO4

  9. Harder examples: Balance the neutralization reaction first: ___H3PO4 + ____ Ca(OH)2 ____ Ca3(PO4)2 + ____H2O a. How many moles of Ca(OH)2 will it take to neutralize 3 moles of H3PO4? b. How many liters of 1.5 M Ca(OH)2 will it take to neutralize 1.2 moles of H3PO4?    c. How many liters of 2.9 M H3PO4 will it take to neutralize 300 mL of 0.7 M Ca(OH)2? 2 3 6 ) = ( 3 mol H3PO4 3 mol Ca(OH)2 4.5 mol Ca(OH)2 2 mol H3PO4 ) = ( 1.2 mol H3PO4 3 mol Ca(OH)2 1.8 mol Ca(OH)2 2 mol H3PO4 1.8 mol Ca(OH)2 ÷ 1.5 M = 1.2 Liters 0.3 L x 0.7 M = 0.21 mol Ca(OH)2 ) = ( 0.21 mol Ca(OH)2 2 mol H3PO4 0.14 mol 3 mol Ca(OH)2 H3PO4 0.14 mol H3PO4 ÷ 2.9 M = 0.05 Liters

  10. End Point The point at which the indicatorchanges color is an end point.

  11. Steps for Titration Calculations • Start by writing a balanced(neutralization) equation (to determine mole to mole ratios). Neutralization reactions are always double displacement reactions. • Write down all of your information underneath the chemical equation. • Determine moles known(Acid/Base). M x L = moles • Use mole to mole ratio to determine moles of unknown(Acid/Base). • Determine molarityof unknown (mol/L)

  12. Thinking problem: Let’s say you had to add 200 mL of 1.0 M NaOH to 200 mL of HCl before the indicator turned pink. What molarity must the acid have? Thinking problem #2: What if you had to add 400 mL of 1.0 M NaOH to 200 mL of HCl before the indicator turned pink. What molarity must the acid have? 1 M HCl 2 M HCl

  13. Titration Problem #1 A 25 mL solution of HCl is completely neutralized by 18 mL of 1.0M NaOH. What is the concentration of the HCl solution? HCl + NaOH  NaCl + H2O x 0.018 L NaOH 0.018 mol NaOH 1.0 mol NaOH = 1 L NaOH 0.018 mol NaOH x 1.0 mol HCl 0.018 mol HCl = 1.0 mol NaOH 0.018 mol HCl 0.72 M = 0.025 L

  14. YOUR Titration Problem 15.0 mL of 1.50 M HCl is titrated with 10.8 mL of NaOH. Determine the molarity of the sodium hydroxide solution. Write the reaction first: HCl + NaOH  NaCl + H2O x 0.015 L HCl 0.0225 mol HCl 1.5 mol HCl = 1 L HCl 0.0225 mol HCl x 1.0 mol NaOH 0.0225 mol NaOH = 1.0 mol HCl 0.0225 mol NaOH 2.08 M = 0.0108 L

  15. And one more… You have a 200 mL sample of NaOH of an unknown concentration. It takes 75 mL of a 1.25 M H2SO4 in order to neutralize the solution. What is the concentration of NaOH? Write the reaction first: H2SO4 + 2 NaOH  Na2SO4 + 2 H2O x 0.075 L H2SO4 0.094 mol H2SO4 1.25 mol H2SO4 = 1 L H2SO4 0.094 mol H2SO4 x 2 mol NaOH 0.188 mol NaOH = 1 mol H2SO4 0.188 mol NaOH 0.94 M = 0.2 L

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