# Solution stoichiometry Description Special case >>>>>> Analysis (stoich) ( ) A.P. Chem. Structured Overview Chapter 4 Making Reacting like dissolves like.

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solution stoichiometry Description Special case >>>>>> Analysis (stoich) ( ) A.P. Chem. Structured Overview Chapter 4 Making Reacting like dissolves like What’s really present (the major species) Electrolytes (Arrhenius) non weak strong Acids and Bases H+H+ OH - Quantify (molarity) Calculations (mol/L) moles mass of solute needed Dilute (M 1 V 1 = M 2 V 2 ) PPT Acid/Base REDOX Molec. (formula) Ionic Net ionic gravimetric volumetric titration Solubility rules oxidation states process agents Balance (both acidic and basic) (M a V a = M b V b ) (MV)

9 multiple choice 4 equation writing 3 problem solving 1 short answer 2 one point bonus questions (based on extra little things) 15M (V) = 0.16M (20000.0 mL) * Free Response How much of a 15 M solution would be needed to make 20.0 L of a 0.16 M sol’n? M 1 V 1 = M 2 V 2 V ≈ 200 mL * No Calculators on M.C. section; must use mental math

The distinctive odor of vinegar is due to acetic acid, HC 2 H 3 O 2. Acetic acid reacts with sodium hydroxide in the following fashion: HC 2 H 3 O 2(aq) + NaOH (aq) → H 2 O (l) + NaC 2 H 3 O 2(aq) Write the net ionic equation. If 2.50 mL of vinegar requires 34.9 mL of 0.0960 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00 qt. sample of vinegar? HC 2 H 3 O 2(aq) + OH - (aq) → H 2 O (l) + C 2 H 3 O 2 - (aq) M a V a = M b V b M a = 0.0960 mol 34.9ml = 1.34 mol L 2.50 mL ()() 1.34 mol 60.0 g 1 L = L 1 mol 1.06 qts )))((( 75.8 g/qt Weak acid

Stibnite, (Sb 2 S 3 ), is the most important ore containing antimony. A sample of the ore was chemically treated to produce antimony (III) ions in solution. The antimony was oxidized to antimony (V) by adding 25.00 mL of 0.0233M KMnO 4 solution. The excess KMnO 4 was titrated with a 0.0843 M Fe 2+ solution; 2.58 mL was required, producing Fe 3+ and Mn 2+ aqueous ions. All reactions were carried out in acidic solutions. Calculate the mass of the Sb in the sample. Sb 3+ Sb 5+ MnO 4 - Sb 5+ Mn 2+ Fe 3+ Mn 2+

Stibnite, (Sb 2 S 3 ), is the most important ore containing antimony. A sample of the ore was chemically treated to produce antimony (III) ions in solution. The antimony was oxidized to antimony (V) by adding 25.00 mL of 0.0233M KMnO 4 solution. The excess KMnO 4 was titrated with a 0.0843 M Fe 2+ solution; 2.58 mL was required, producing Fe 3+ and Mn 2+ aqueous ions. All reactions were carried out in acidic solutions. Calculate the mass of the Sb in the sample. 8H + + MnO 4 - + 5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ 2.58 mL Fe 2+ 0.0843 mol 1 mol MnO 4 - = 1000 mL 5 mol Fe 2+ )))((( 0.0000435 mol MnO 4 - (extra) ( ()) 25.00 mL MnO 4 - 0.0233 mol = 1000 mL 0.000583mol MnO 4 - (total) 0.000583mol (total) -0.0000435 mol (extra) 0.000540 mol MnO 4 -(used) 16H + + 2MnO 4 - + 5Sb 3+ → 2Mn 2+ + 8H 2 O + 5Sb 5+ ( () ) 0.000540 mol MnO 4 - 5 mol Sb 3+ = 2 mol MnO 4 - 0.00135 mol Sb 3+ (ore) 0.00135 mol Sb 3+ () () 121.8 g Sb 3+ 1 mol Sb 3+ = 0.164 g Sb 3+ (ore)

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