5/19/14 QOTD 1. Have your HW out to be checked off (p. 70) 2. Please turn in your “Chemistry In My Life” posters on the front counter! Pick up: answer key to the HW (p.70) and notes sheet for neutralization and titration 3. QOTD: pH = - log [H+] pOH = - log[OH-] pH + pOH = 14 a. What is the pH of a 0.004 M HCl solution? b. What is the pH of a 0.07 M NaOH solution?

QOTD Answer pH = - log[H+] = - log(0.004) = 2.4 This is a base so you need to find pOH first! pOH = - log[OH-] = - log (0.07) = 1.15 pH = 14 – pOH = 14 – 1.15 = 12.85

Learning Target You should be able to describe and write what happens when an acid is added to a base. Quiz tomorrow on acids, bases, and pH (does NOT include today’s material  )

Neutralization and Titrations NOTES TIME!

Neutralization Reactions Reactions in which an acid and a base react in an aqueous solution to produce a salt and water. HCl + NaOH  NaCl + H2O What types of reactions are these? Double Displacement

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

YOUR TURN: Complete & Balance the reactions 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

TITRATIONS! http://www.youtube.com/watch?v=ZX4mEkjIuUc

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

Equivalence Point When the number of moles of hydrogen ions equals the number of moles of hydroxide ions. In other words, when you have the same mole ratio of reactants in the balanced chemical equation. HCl + NaOH  NaCl + H2O Need a 1:1 ratio of acid to base H2SO4 + 2 NaOH  Na2SO4 + 2H2O Need a 1:2 ratio of acid to base 2HCl + Ca(OH)2  CaCl2 + 2H2O Need a 2:1 ratio of acid to base

Example: How many moles of HCl will it take to neutralize 6 moles of NaOH? HCl + NaOH  H2O + NaCl 1:1 mole ratio so 6 moles of HCl react with 6 moles NaOH

YOUR TURN! (work with your neighbor!) Do the simple examples a-d. We will go over them in about 3 minutes.

YOUR TURN! Answers 2 HCl + Ca(OH)2  2 H2O + CaCl2 12 moles HCl How many moles of HCl would it take to neutralize 6 moles of Ca(OH)2?   b. How many moles of NaOH would it take to neutralize 6 moles of H2SO4? 2 HCl + Ca(OH)2  2 H2O + CaCl2 12 moles HCl H2SO4 + 2 NaOH  2 H2O + Na2SO4 12 moles NaOH

( ) = ( ) = 9 mol Ca(OH)2 2 mol H3PO4 6 mol H3PO4 3 mol Ca(OH)2   c. How many moles of H3PO4 would it take to neutralize 9 moles of Ca(OH)2? d. How many moles of H3PO4 would it take to neutralize 6.4 moles of Ca(OH)2? 2 H3PO4 + 3 Ca(OH)2  3 H2O + Ca3(PO4)2 ( ) = 9 mol Ca(OH)2 2 mol H3PO4 6 mol H3PO4 3 mol Ca(OH)2 2 H3PO4 + 3 Ca(OH)2  6 H2O + Ca3(PO4)2 ( ) = 6.4 mol Ca(OH)2 2 mol H3PO4 4.3 mol H3PO4 3 mol Ca(OH)2

___H3PO4 + ____ Ca(OH)2  ____ Ca3(PO4)2 + ____H2O Your turn again! Harder problems! Answers in about 5 minutes Harder problems: 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?   2 3 1 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

___H3PO4 + ____ Ca(OH)2  ____ Ca3(PO4)2 + ____H2O Harder problems: Balance the neutralization reaction first: ___H3PO4 + ____ Ca(OH)2  ____ Ca3(PO4)2 + ____H2O    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 1 6 ( ) = 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

End Point The point at which the indicator changes color is an end point.

Titrating an acid Volume of base added We get this curve because pH is logarithmic!

Turn and talk: Why do you think we use phenolphthalein if turns color AFTER the equivalence point? Faint pink at pH = 8.3 (not 7!) pH Look at how fast pH goes up! It only takes a drop or two to go from the equivalence point to pink! Volume of base added 1 drop is about 0.05 mL  very small error

Steps for Titration Calculations Start by writing a balanced (neutralization) equation (to determine mole to mole ratios). Neutralization reactions are always double displacement reactions. One of the products is ALWAYS water! Write down all of your information underneath the chemical equation. Determine moles known acid (or base) M x L = moles Use mole to mole ratio to determine moles of unknown base (or acid). Determine molarity of unknown (mol/L)

1 M HCl 2 M HCl TURN AND TALK: 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

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

YOUR Titration Problem (work with your neighbor) 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 0.015 L HCl x 1.5 mol HCl = 0.0225 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

And one more… H2SO4 + 2 NaOH  Na2SO4 + 2 H2O 0.075 L H2SO4 x 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 0.075 L H2SO4 x 1.25 mol H2SO4 = 0.094 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

Let’s do a real titration! Draw a picture of the setup and write the chemical equation. Once we get the volume of base added (from the buret) we can calculate the molarity of the acid. Formula of Base  NaOH Formula of Acid  HCl Molarity of Base  0.2 M NaOH Amount of acid in flask  15.0 mL Initial Buret reading   Final Buret reading Amount of base added (take final buret – initial buret)

HOMEWORK: OPTIONAL practice quiz