# Solution Concentration Molarity = mol solute = M L sol’n Most commonly used unit of concentration A 0.50 M sol'n = 0.50 mol solute 1.0 L sol'n Use it as.

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Solution Concentration Molarity = mol solute = M L sol’n Most commonly used unit of concentration A 0.50 M sol'n = 0.50 mol solute 1.0 L sol'n Use it as a conversion factor – Can change 1 - moles (or grams) of solute to L of sol'n 2 - L of sol'n to mol of solute – Which can be converted to grams

Making Solutions What is the molarity of a solution of 29.25 g of NaCl in 500 mL of water solution? – Add water to the solid to the desired volume. – Solutions volumes are NOT additive

Types of problems 1. Deter M given the amount (mol or g) of solute and volume of solution Convert g mol and ÷ by L sol'n 2. Deter the amt of solute - in grams or moles -in a given vol of sol'n (L x M = moles) 0.50 M means 0.50 mol/1L use M as conver. fact. to convert L mol Convert moles g if necessary 3.Determine the vol of sol’n containing a given amt of solute - in moles or grams If given g convert to mol Use M convert mol vol of sol'n in L

1. Calc the conc. (M) of a solution created by dissolving 0.60 mol of NaOH (molar mass = 40.0 g/mol) in enough water to make 1.75 L of solution.

2. Determine the number of grams of CaCO 3 (mm = 100.1 g/mol) in 3.55 L of a 1.5 M solution

3. How many liters of a 2.50 M solution of H 2 SO 4 (molar mass = 98.0 g/mol) could be created from 27.0 g of H 2 SO 4 ?

Dilutions Problems Dilution means to add more solvent, usually water, and reduce the solution’s conc (M). – M x L = moles Molarity x vol (L) = moles solute – Mole solute before = mole solute after – M b V b = M a V a M is not a conversion factor in these problems

Dilutions Problems A 1.685 L solution of HCl is 0.055 M. If the solution is diluted to 2.500 L what is the new molarity? How many mL of a 0.55 M solution could be made from 1.50 L of a 1.00 M stock solution?

What mass of solid aluminum hydroxide (mm = 78.0 g/mol) is produced when 50.0 mL of 0.200 M Al(NO 3 ) 3 (mm = 213.0 g/mol) is added to 200.0 mL of 0.100 M KOH (mm = 56.1 g/mol)? Al(NO 3 ) 3 (aq) + 3KOH (aq)  Al(OH) 3 (s) + 3KNO 3 (aq)

Solution Stoich What mass, in grams, of AgCl will precipitate when 0.050 L of a 0.0500 M solution of AgNO 3 reacts with 25.0 mL of.0330 M NaCl ? AgNO 3 (aq) + NaCl (aq)  NaNO 3 (aq) + AgCl (s) vol A  mol A  mol B  g B

Electrolytes Electrolyte Substance that, when dissolved in water, produces an solution that conducts an electric current. 2 requirements 1-Charged particles 2-Mobility Ionic compounds and acids (only molecular electrolytes)

Electrolytes Salts - made up of charged particles NaCl (s)  Na + (aq) + Cl - (aq) CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq) Molecules are usually nonelectrolytes Acids = molecular electrolytes HCl (g)  H + (aq) + Cl - (aq) H 2 SO 4 + (g)  H + (aq) + HSO 4 - (aq)

Nonelectrolytes Nonelectrolyte compound that, when dissolved in water, produces aq solution that does NOT conduct an electric current molecular compounds – C 6 H 12 O 6 (s)  C 6 H 12 O 6 (aq) no charged particles present

Strong versus Weak Electrolytes Strong electrolytes – Usually ionize completely or almost completely – Produce solutions that conduct a strong current Weak electrolytes – Ionize on the order of 1-10% – Produce solutions that conduct a weak current

Precipitation Reactions Why things dissolve Precipitate = solid ionic compound formed when water solutions of 2 diff ionic compounds are mixed Know rules fig 4.3 p 78 Know what ions combine to form a ppt

Net Ionic Equations NIE show – 1. Balance the formula equation – 2. Dissociate (pull apart) all (aq) compounds Subscripts for R and P become coefficients CaCl 2(aq) Ca 2+ (aq) + 2Cl - (aq) – 3. Remove all spectator ions spectator ions do nothing in the reaction and are excluded from the equation NaOH (aq) + Cu(NO 3 ) 2(aq) --> NaNO 3(aq) + Cu(OH) 2(s)

First, balance the equation 2NaOH (aq) + Cu(NO 3 ) 2(aq) --> 2NaNO 3(aq) + Cu(OH) 2(s) Write ionic eq (dissociate all aq compounds) and cancel the spectator ions 2Na + + 2OH - + Cu 2+ + 2NO 3 - --> 2Na + + 2NO 3 - + Cu(OH) 2(s) Write NIE (write (+) ion first) Cu 2+ (aq) + 2OH - (aq) ---> Cu(OH) 2(s) This is all that really happened!!!

Net Ionic Equations – BaCl 2 and Na 2 SO 4 are mixed. Write NIE for the reaction if a precipitate forms. Ions present are Ba 2+ Cl - Ag + SO 4 2- – Only + and - ions join together, therefore – Possible ppts are (Ba 2 + and SO 4 2- ) or (Na + and Cl - ) » All chlorides are soluble (ex Ag + and any Pb ion) » All sulfates are soluble (ex Ba 2+ ) – Barium and sulfate ions will form a ppt Ba 2+ + 2Cl - + 2Na + + SO 4 2- BaSO 4 (s) + 2Na + + 2Cl - Ba + (aq) + SO 4 2- (aq) BaSO 4 (s)

NIE from Reactants Write NIE for any reaction (or no rx) that occurs when (aq) solutions of NaCl (aq) + AgNO 3(aq) are mixed Na + + Cl - + Ag + + NO 3 - --> possible ppt = AgCl or NaNO 3 – All nitrate are soluble – Most Ag compounds are insoluble therefore ppt is – Ag + (aq) + Cl - (aq) AgCl(s)

Review Write NIE for any rx between aq sol’n of – AgNO 3 (aq) + Ca(OH) 2 (aq) – AlCl 3 (aq) + Pb(NO 3 ) 2 (aq)

How many grams of AgNO 3 (ag) are produced when 0.85L of 2.00 M HNO 3 solution is added to 216 g of Ag according to the equation. Which reactant is the limiting reactant? 3 Ag(s) + 4 HNO 3 (aq)  3 AgNO 3 (aq) + NO(g) + 2 H 2 O(l)

Acids Commercially important Contain H and a nonmetal or polyatomic ion Taste sour – found in many foods Rx with many other substances – w/ metals to produce H 2 gas – w/ carbonates to produce CO 2 and a salt Molecular electrolytes – Produce H + ions in water solution Neutralize bases

Bases Produce OH - in water solution Feel slippery Taste bitter Commercially important Present in many cleaning agents Usually metal hydroxides, ammonia or amines, and some anions Neutralize acids

Acids – Proton donors Acid produce H + ions – 2 types – strong acids and weak acids Know examples in text pg 83 Strong acids – ionize (break apart) 100% – HCl(aq) H + (aq) + Cl - (aq) – Acids lose only 1 H + in aq sol’n – H 2 SO 4 (aq) H + (aq) + HSO 4 - (aq) Weak acids - ionize only slightly – less than 5% – Use a 2 headed arrow to show both reactants and products are present – HF(aq) H + (aq) + F - (aq)

Acids in Water Strong acids: Ionize 1 proton completely initial 1000 0 0 H 2 SO 4(l) H + (aq) + HSO 4 - (aq) final 0 1000 1000 Weak acids : Ionize 1 proton in eq system initial 1000 0 0 H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) final 995 5 5

Bases – Proton acceptor Bases produce OH - ions in aq solution – Bases are proton grabbers Strong and weak bases (pg83) – Strong bases = grp 1 & 2 metal hydroxides – They dissociate all of their hydroxide ions NaOH (s) Na + + OH - Ca(OH) 2(s) Ca 2+ + 2OH - – Weak bases (H + grabbers) do not contain OH - They react with water (by grabbing a proton) leaving OH - – H 2 O as a reactant and use means reversible rx – NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH - (aq) – F - (aq) + H 2 O(l) HF(aq) + OH - (aq)

Bases in water Strong Bases loses all their OH - to water Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) Weak bases break H 2 O apart and form OH - ions (grabs a proton) F - (aq) + H 2 O (l) HF (aq) + OH - (aq)

Acids and Bases in H 2 O Write NIE for reactions, in H 2 O, of – SA SBWA SB – SA WBWA WB Write SA & SB as the active ion – H + or OH - Write WA & WB as the entire molecule – HF CH 3 COOH NH 3 CH 3 NH 2 H + ion jumps to the base

Acids & Bases in H 2 O SA= HCl (g) H + (aq) + Cl - (aq) – Strong acids lose only 1 proton (i.e. H 2 SO 4 ) WA=H 3 PO 4 (g) H + (g) + H 2 PO 4 - (g) – Weak acid lose only 1 proton and set up an equilibrium system use a 2 headed arrow SB= Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) – All OH - ions are removed from the molecule WB= NH 3(g) + H 2 O (l) NH 4 + (aq) + OH - (aq) – WB grab H + from H 2 O to produce OH - ions

Know Rx for Strong and Weak Acids and Bases in Water HCl (aq) H + (aq) + Cl - (aq) H 2 CO 3 (aq) H + (aq) + HCO 3 - (aq) Ca(OH) 2 (s) Ca 2+ (aq) + 2OH - (aq) NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq)

Acids & Bases in Reactions Represent each of the following in reactions, or water solution: – HNO 3 H 2 SO 4 HF HC 2 H 3 O 2 H 3 PO 4 – H + H + HF HC 2 H 3 O 2 H 3 PO 4 – KOH Ca(OH) 2 NH 3 F- CH 3 NH 2 – OH - OH - NH 3 F- CH 3 NH 2

Acid Base Neutralization Reactions Acids and bases neutralize each other All H + from the acid and all OH - from the base react to produce HOH (H 2 O) In NIE reactions: – Strong acids are represented by the H + ion – Strong bases are represented by the OH- ion – Weak acid and bases are represented by the entire molecule

Strong Acid + Strong Base Active ions are H + and OH - – They combine to form water Write the equation for the reaction between nitric acid and potassium hydroxide – HNO 3(aq) + KOH (aq) H 2 O (l) + KNO 3(aq) Break apart (aq) formulas and cancel spectators H + + NO 3 - + K + + OH - H 2 O (l) + K + + NO 3 - – H + (aq) + OH - (aq) H 2 O(l) – In titrations use formula equations

Strong Acid Weak Base Represent the strong acid as H + and show the entire molecular formula of the weak base – Combine the H + with the formula for the base Write the eq for the reaction of nitric acid, HNO 3, and ammonia, NH 3. – HNO 3(aq) + NH 3 (aq) NH 4 NO 3 (aq) Break apart (aq) formulas (not the weak base) and cancel spectators The weak base grabs H + from the acid – H + (aq) + NO 3 - (aq) + NH 3(aq) NH 4 + (aq) + NO 3 - (aq) – H + (aq) + NH 3(aq) NH 4 + (aq)

Weak Acid and Strong Base Show the entire formula for the acid, and OH - for the base. – The H + from the acid and OH - from the base form water, and the ion from the acid remains Write the equation for the reaction of carbonic acid and potassium hydroxide – H 2 CO 3(aq) + 2KOH (aq) 2H 2 O (l) + K 2 CO 3 (aq) – H 2 CO 3(aq) + 2K + (aq) + 2OH - (aq) 2H 2 O (l) + 2K + (aq) + 2CO 3 2- (aq) Break apart (aq) species (not the WA) and cancel spectators – H 2 CO 3(aq) + 2OH - (aq) 2H 2 O (l) + CO 3 2- (aq)

A & B NIE Neutralization Equations SA SB Nitric acid + sodium hydroxide – H + (aq) + OH - (aq) H 2 O (l) SA WB Hydrochloric acid + ammonia – H + (aq) + NH 3(g) NH 4 + (aq) WA SB Carbonic acid + potassium hydroxide – H 2 CO 3(aq) + OH - (aq) H 2 O (l) + HCO 3 - (aq)

A & B equations Hydrobromic acid + potassium hydroxide Nitric acid + ammonia Phosphoric acid + lithium hydroxide

Acid Base Formula Eq Neutralization Reactions are double displacement reactions – HA + MOH  HOH (water) + MA (a salt) – HA + NH 3  NH 4 A (an ammonium salt) – HCl + LiOH  – HCl + LiOH  H 2 O + LiCl – H 2 CO 3 + NH 3  – H 2 CO 3 + 2NH 3  (NH 4 ) 2 CO 3 – HF + Al(OH) 3  – 3HF + Al(OH) 3  3H 2 O + AlF 3

Acid Base Titration Lab procedure used to determine the molarity of a solution. You completely reacts 2 solution. – React a solution of known conc with a solution of unknown conc – When equal moles of the 2 solutes have completely reacted with one another, the reaction ends End point (or equivalence point) of the titration is determined by a color change in the solution At the end point: moles solute A = moles solute B

Acid Base Titration React a sol'n of known M w/ a sol'n of unknown M – Solution of known conc is called standard sol’n – Determine the vol of each sol'n used in the titration – you know the conc of one sol'n, but not the other – You’ll know the vol of both solutions – At end point of the titration = number of moles of each solute have been reacted mol solute knwn = mol solute unk You can calc the # mol of solute in the known solution from its vol and M – # mol solute knwn = vol knwn (L) x M knwn = moles knwn = mol unk M unk = mol unk/L unk sol'n

Acid Base Titration 1. Use L and M of the known (standard) solution to find the moles in the known solution 2. Use stoich factor to calc mol of unknown in it’s solution 3. Find M of the unknown M = mol unk L unk

Beyond A/B Titration Math formula Write a balanced formula equation (NOT NIE) - 1 - Calc moles of the known substance - 2 - Convert to moles of the unknown with the stoich factor - 3 - Divide by the L of the unknown to produce Molarity 2. Calc # moles of solute in the known sol'n mol kwn = M kwn x vol (L) kwn 3. Calc the # moles of solute in the unk sol'n mol unkwn = mol kwn x mol unkwn (the stoich factor, mol kwn from the bal eq) 4. Divide the moles of solute in the unknown solution by the # of liters of solution used in the titration M unkwn = mol unkwn L unkwn

Titration Math a A 15.0 mL sample of HCL is titrated to the eq pt with 25.0 mL of 0.500 M NaOH. Calc the M of the acid. 1. HCl(aq) + NaOH(aq) H 2 O(l) The stoich factor is 1:1 2. n NaOH = 0.025 L NaOH x 0.500 mol NaOH = 0.0125 mol NaOH 1 L NaOH 3. Use the stoich factor to deter moles H + n HCl = 0.0125 mol NaOH x 1 mol HCl = 0.0125 mol HCl 1 mol NaOH 4. M H+ = 0.0125 mol HCl = 0.833 M HCl 0.0150 L HCl

Titration Math b A 15.0 mL sample of oxalic acid, H 2 C 2 O 4 is titrated to the eq pt with 25.0mL of 0.500 M NaOH. Calc the M of the acid. 1. H 2 C 2 O 4 + 2NaOH 2H 2 O + Na 2 C 2 O 4 » Stoic factor is 1 mol H 2 C 2 O 4 = 2 mol NaOH 2. n OH- = 0.025 L NaOH x 0.500 mol NaOH = 0.0125 mol NaOH 1 L OH- 3. n ox = 0.0125 mol NaOH x 1 mol ox = 0.00625 mol ox 2 mol NaOH 4. M H2C2O4 = 0.00625 mol ox = 0.417 M Ox acid 0.0150 L ox

Acids and Bases Definitions – Strong vs weak Equations of acids and bases in water – SA, WA, SB, WB Acid/Base neutralization reactions – SA/SB, WA/SB, SA/WB A/B Titration math

Oxidation Reduction Reactions Involve a transfer of e- – One reactant loses e-s and is oxidized – One reactant gains e-s and is reduced Oxidation and reduction occur together – e-s (charge) and atoms are conserved Oxidizing agent = subst reduced Reducing agent = subst oxidized – Cu 2+ (aq) + Zn 0 (s) Cu 0 (s) + Zn 2+ (aq) – Zn 0 (s) + 2H + (aq) Zn 2+ (aq) + + H 2 (g)

Oxidation Numbers Way to keep track of e-s in redox rx In molecules and PAI the oxid # are = to number of e-s shared between atoms Know rules for assigning oxid #s p89 Oxidation = increase in oxid # – Loss of electron(s) Reduction = decrease in oxid # – Gain of electron(s)

Half reactions A redox rx can be divided into an oxidation and a reduction ½ reactions – Cu 2+ (aq) + Zn 0 (s) Cu 0 (s) + Zn 2+ (aq) – Oxidation ½ rx = Zn 0 (s) Zn 2+ (aq) + 2 e- – Oxidation number of Zn goes up – Zn is oxidized and is the reducing agent » Zn losses e-s – Reduction ½ rx = Cu 2+ (aq) + 2e- Cu 0 (s) – Oxidation number of Cu 2+ goes down – Cu 2+ is reduced and is the oxidizing agent » Cu 2+ gains f e-s

Balancing Redox Equations All redox equations must be balanced for both atoms and charge Know steps in the process Balance all equation on the worksheet in acidic and basic solutions Redox reactions can be used for a titration reaction – Follow steps for any titration

Balancing Redox Equations Bal all atoms other than H and O Bal O using water Bal H using H + ions Bal charges by placing e - ’s on the side with more positive charges – Charge, NOT oxidation number Bal both ½ rxs so e - s gained = e - s lost Combine ½ rx equations and cancel If basic, add OH- to each side to cancel out all H + ions present – cancel H 2 O’s

Cr 2 O 7 2- + NO 2  Cr 3+ + NO 3 1- Cr 2 O 7 2-  Cr 3+ NO 2  NO 3 - Cr 2 O 7 2-  2Cr 3+ NO 2  NO 3 - Cr 2 O 7 2-  2Cr 3+ + 7H 2 O H 2 0 + NO 2  NO 3 - 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O H 2 0 + NO 2  NO 3 - + 2H + 14H + + Cr 2 O 7 2- + 6e -  2Cr 3+ + 7H 2 O (x6) H 2 0 + NO 2  NO 3 - + 2H + + 1e - 14H + + Cr 2 O 7 2- + 6e -  2Cr 3+ + 7H 2 O 6H 2 0 + 6NO 2  6NO 3 - + 12H + + 6e - 14H + + Cr 2 O 7 2- + 6H 2 0 + 6NO 2  2Cr 3+ + 7H 2 O + 6NO 3 - + 12H + 2H + + Cr 2 O 7 2- + 6NO 2  2Cr 3+ + H 2 O + 6NO 3 -

Weak Acid + Weak Base Represent both acid and base as entire formula. The acid loses 1 H + to the base CH 3 COOH (l) + NH 3 (g) NH 4 + (aq) + CH 3 COO - (aq) H 2 CO 3(aq) + CH 3 NH 2(aq) HCO 3 - (aq) + CH 3 NH 3 + (aq)

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