SOLUTION SOLUTION STOICHIOMETRY LECTURE STOICHIOMETRY LECTURE

33

ACIDS AND BASESACIDS AND BASES

ArrheniusArrhenius

AcidsAcids Give off HGive off H++

BasesBases Give off OHGive off OH--

Substances that caused a problem???Substances that caused a problem??? Main example – NHMain example – NH33

BrBrønsted-Lowryønsted-Lowry

Came up with new definition to Came up with new definition to account for theseaccount for these

AcidsAcids Give off (donate) protons (HGive off (donate) protons (H++))

BasesBases Accept protons (HAccept protons (H++))

Remember conjugates??Remember conjugates??

NeutralizationNeutralization

A special type of double replacement A special type of double replacement reactionreaction Reactants =Reactants =

AcidAcid BaseBase

ProductsProducts WaterWater A salt (ionic compound)A salt (ionic compound)

NeutralizationNeutralization

Strong Acid + Strong BaseStrong Acid + Strong Base

Hydrochloric acidHydrochloric acid(aq)(aq) + sodium hydroxide + sodium hydroxide(aq)(aq)

DemoDemo MolecularMolecular

HCl HCl (aq)(aq) + NaOH + NaOH (aq)(aq) --> NaCl --> NaCl (aq)(aq) + HOH + HOH (l)(l)

Complete IonicComplete Ionic HH++

(aq)(aq) + Cl + Cl--(aq)(aq) + Na + Na++

(aq)(aq) + OH + OH--(aq) (aq) --> Na--> Na++

(aq)(aq) + Cl + Cl--(aq)(aq) + HOH + HOH(l)(l)

Net IonicNet Ionic HH++

(aq)(aq) + OH + OH--(aq)(aq) --> HOH --> HOH(l)(l)

Formation of water

Weak Acid + Strong BaseWeak Acid + Strong Base

Acetic acid Acetic acid (aq)(aq) + potassium hydroxide + potassium hydroxide (aq)(aq)

MolecularMolecular HCHC22HH33OO2 (aq)2 (aq) + KOH + KOH (aq) (aq) --> KC--> KC22HH33OO2 (aq) 2 (aq) + HOH + HOH (l)(l)

Complete IonicComplete Ionic HCHC22HH33OO2 (aq)2 (aq) + K + K++

(aq)(aq) + OH + OH--(aq)(aq) --> K --> K++

(aq)(aq) + C + C22HH33OO22--(aq)(aq) + +

HOHHOH(l)(l)

Net IonicNet Ionic HCHC22HH33OO2 (aq)2 (aq) + OH + OH--

(aq)(aq) --> C --> C22HH33OO22--(aq)(aq) + HOH + HOH(l)(l)

StoichiometryStoichiometry

1.1. Write equations.Write equations.2.2. Balance net ionic equation.Balance net ionic equation.3.3. Find moles (from original solutions).Find moles (from original solutions).4.4. Limiting reactantLimiting reactant5.5. Mole ratioMole ratio6.6. Get it out of molesGet it out of moles

Example -Example - How many grams of water are formed when 50.0 mL How many grams of water are formed when 50.0 mL

of 0.356 M HI react with 28.0 mL of 0.488 M barium of 0.356 M HI react with 28.0 mL of 0.488 M barium hydroxide?hydroxide?

TitrationTitration Delivery with a buret of a known solution (Delivery with a buret of a known solution (titranttitrant) )

into an unknown solution (into an unknown solution (titratetitrate or or analyteanalyte) until ) until the the endpointendpoint or or equivalence pointequivalence point VocabularyVocabulary

Titrant = solution of known concentration, in buretTitrant = solution of known concentration, in buret Titrate = solution of unknown concentration, in beakerTitrate = solution of unknown concentration, in beaker Endpoint = point where an indicator first changes color, Endpoint = point where an indicator first changes color,

titrant exactly reacted with the titratetitrant exactly reacted with the titrate Equivalence point = HUGE!!!Equivalence point = HUGE!!!

Demo = iron ions and permanganate (we’ll do Demo = iron ions and permanganate (we’ll do it for real later in the year & simulated)it for real later in the year & simulated)

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/redoxNew/redox.html

HUGE POINTHUGE POINT

Equivalence pointEquivalence point Occurs when Occurs when MOLES = MOLES MOLES = MOLES

(stoichiometrically)(stoichiometrically)

Really important for titration calculations!!Really important for titration calculations!!

Example Titration Problem Example Titration Problem from AP examfrom AP exam

0.2640 g of sodium oxalate is dissolved in a flask 0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium permanganate and requires 30.74 mL of potassium permanganate (from a buret) to titrate it and cause it to turn pink (from a buret) to titrate it and cause it to turn pink (the end point). (the end point).

The equation for this reaction is: The equation for this reaction is: 5 Na5 Na22CC22OO4 (aq)4 (aq) + 2 KMnO + 2 KMnO4 (aq)4 (aq) + 8 H + 8 H22SOSO4 (aq)4 (aq)

2 MnSO2 MnSO4 (aq)4 (aq) + K + K22SOSO4 (aq)4 (aq) + 5 Na + 5 Na22SOSO4 (aq)4 (aq) + 10 CO + 10 CO2 (g)2 (g) + 8 H + 8 H22O O (l)(l)

(a) How many moles of sodium oxalate are present in the (a) How many moles of sodium oxalate are present in the flask? flask? (b) How many moles of potassium permanganate have (b) How many moles of potassium permanganate have been titrated into the flask to reach the end point? been titrated into the flask to reach the end point? (c) What is the molarity of the potassium (c) What is the molarity of the potassium permanganate?permanganate?