CHEM 101 LECTURE NOTES Fall 2003 Dr. Joy Heising S Chapter 11 lecture notes Ch. 11 Aqueous solution reactions

Size: px

Start display at page:

Download "CHEM 101 LECTURE NOTES Fall 2003 Dr. Joy Heising S Chapter 11 lecture notes Ch. 11 Aqueous solution reactions"

Magdalen Lindsey
5 years ago
Views:

1 Ch. 11 Aqueous solution reactions Ch. 3 review: Molarity = moles solute Liters solution If 500. ml of a 2.80 M solution of NaOH is added to 75.0 ml of a 3.68 M solution of H 3 PO 4, the resulting solution is M in Na 3 PO 4 and M in. 1) write a balanced equation: H 3 PO 4 (aq) + 3NaOH (aq) Na 3 PO 4 (aq) + 3H 2 O (l) 2) figure out # moles of H 3 PO 4 /NaOH involved in reaction L x 3.68 M = mol H 3 PO 4 1 L L x 2.80 M = 1.40 mol NaOH 1 L 3) set up a mole ratio to determine limiting reacting 3 mol NaOH = 3 (from equation) 1.40 mol NaOH = mol H 3 PO mol H 3 PO 4 tiny on top big on bottom H 3 PO 4 is limiting reactant 4) solve for amount of Na 3 PO 4 generated from limiting reactant and calculate molarity mol H 3 PO 4 x 1 mol Na 3 PO 4 x. 1. = M Na 3 PO 4 1 mol H 3 PO 4 (0.575 L) we re half done! 5) solve for amount of NaOH left & calculate molarity (1.40 mol NaOH [0.276 mol H 3 PO 4 x 3 NaOH]) x. 1. = M NaOH 1 H 3 PO L

2 acid/base titrations indicator a compound that exhibits different colors in solutions of different acidities. Equivalence point chemically equivalent amounts of reactants have reacted Choose indicator which changes color close to equivalence point End point when indicator changes color & titration is stopped Standard solutions solutions of accurately known concentrations Standardization process by which the concentration of a solution is accurately determined by titrating against a primary standard (primary standard) high purity substance which reacts extremely reliably with substance that we wish to standardize. Usually purchased from chemical company. ex. For acids, Na 2 CO 3 H 2 SO 4 (aq) + Na 2 CO 3 (aq) Na 2 SO 4 (aq) + CO 2 (g) + H 2 O (l)

3 1 mole of base doesn t always neutralize 1 mole acid (polyprotic acids, Ba(OH) 2, etc.) convenient sometimes to use an alternate method for expressing concentration that permits a 1:1 (neutralization) ratio between acids and bases. 1 equivalent of an acid = quantity of the acid that will deliver 1 mole H + ions (eq.) similar to moles for HCl, 1 mol = 1 eq HCl H + + Cl for H 2 SO 4, 1 mol = 2 eq H 2 SO 4 2 H + + SO 4 or 0.5 mol = 1 eq 1 mol HCl x 1 mol H + = 1eq. HCl 1 mol H 2 SO 4 x 2 mol H + = 2 eq. H 2 SO 4. 1 mol HCl 1 mol H 2 SO 4 1 equivalent of a base = quantity of base that will deliver 1 mole OH ions (eq.) equivalent weight mass acid(base)/equivalent acid(base) similar to molar mass H 2 SO 4 = g/mol H 2 SO 4 = g/2 eq = g/eq Also have new concentration unit: Normality = # equivalents (acid or base) similar to molarity Liter of solution 1.00 M H 2 SO 4 = 2.00 N H 2 SO M H 2 SO 4 = 1.00 N H 2 SO 4 ex. calculate molarity & normality of a solution containing g H 3 PO 4 in 1.80 L. Molarity g x 1 mol x. 1. = M H 3 PO g 1.80 L mol H 3 PO 4 x 3 eq. H 3 PO 4 = 2.62 N H 3 PO 4 1 L solution 1 mol H 3 PO 4 ex. calculate # equivalents in 0.300g Cu(OH) 2 (s). mm = 97.6 g/mol 0.300g Cu(OH) 2 x 1 mol Cu(OH) 2 x 2 eq Cu(OH) 2 = 6.15 x 10 3 eq g 1 mol Cu(OH) 2 equivalent mass? 97.6/2 = 48.8 g/eq. in a titration, at the equivalence point # equivalents acid = # equivalents of base

9 Dr. Heising s handout on balancing redox reactions Change in Oxidation Number Method 1. Write as much of unbalanced equation as possible. 2. Assign oxidation numbers & identify species undergoing changes in oxidation numbers. 3. Connect pairs of atoms (with changing oxidation numbers) on opposite sides of equation with brackets (or write a separate mini equation). Balance the oxidized/reduced species within the brackets. 4. Write the total # e gained/lost next to bracket. Include subscripts, coefficients, etc, by multiplying. 5. Make the electrons gained/lost equal to one another by multiplying bracketed species with appropriate coefficients. 6. a. balance remaining atoms by inspection b. balance charge on each side of equation. Half Reaction Method 1. Write as much of unbalanced equation as possible. 2. Assign oxidation numbers & construct unbalanced half reactions for oxidized & reduced species. 3. Balance the half reaction atoms 4. Balance the half reaction charges (by adding electrons). 5. Balance the electron transfer (# e gained/lost). 6. Add the half reactions together and cancel like terms. (electrons MUST cancel.)

14 Balance the following in acidic solution by the ½ reaction method: MnO 4 + SO 3 Mn 2+ + SO MnO 4 + SO 3 Mn 2+ + SO 4 split into ½ reactions, balance charge then peripheral atoms MnO 4 + 5e Mn add 8H + to left, balance charge 8H + + MnO 4 8H + + MnO 4 + 5e Mn 2+ add 4H 2 O to right to balance H/check O + 5e Mn H 2 O balanced + 2e 4 add 2H + to right, balance charge SO 3 SO 4 SO 3 H 2 O + SO 3 SO 4 SO 4 + 2e + 2H + add 1H 2 O to left, balance, H/check O + 2e + 2H + balanced 8H + + MnO 4 H 2 O + SO 3 + 5e Mn H 2 O reduction electrons must cancel SO 4 + 2e + 2H + oxidation 2[8H + + MnO 4 5[H 2 O + SO 3 + 5e Mn H 2 O] add together SO 4 + 2e + 2H + ] 5H 2 O + 5SO H + + 2MnO 4 5H 2 O + 5SO H + + 2MnO e 2Mn H 2 O + 5SO e + 10H e 2Mn H 2 O + 5SO e + 10H + cancel like terms 5SO 3 + 6H + + 2MnO 4 2Mn H 2 O + 5SO 4

15 Ex. chlorine gas is bubbled through basic solution to form chlorate ion and chloride ion. Write unbalanced equation: Cl 2 ClO 3 + Cl split into ½ rxns; balance ox/red atoms Cl 2 2ClO e 0 1 add 12OH to left side; add 6H 2 O to right to balance H 12OH + Cl 2 2ClO e + 6H 2 O balanced Cl 2 + 2e 2Cl balanced 12OH + Cl 2 2ClO e + 6H 2 O multiply by coefficients to cancel es 5[Cl 2 + 2e 2Cl ] then add together 12OH + Cl 2 + 5Cl e 2ClO e + 6H 2 O + 10Cl consolidate/cancel 12OH + 6Cl 2 2ClO 3 + 6H 2 O + 10Cl balanced next page balanced by CON method

TYPES OF CHEMICAL REACTIONS

TYPES OF CHEMICAL REACTIONS Precipitation Reactions Compounds Soluble Ionic Compounds 1. Group 1A cations and NH 4 + 2. Nitrates (NO 3 ) Acetates (CH 3 COO ) Chlorates (ClO 3 ) Perchlorates (ClO 4 ) Solubility