MOLARITY = (moles solute) / (vol.solution in liter units)

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1 CHEM 101/105 Stoichiometry, as applied to Aqueous Solutions containing Ionic Solutes Lect-05 MOLES - a quantity of substance. Quantities of substances can be expressed as masses, as numbers, or as moles. It is equally informative to make reference to.578 grams of sulfur, 4.17 E- molecules of methane, or 0.10 moles of fluorine. For the moment dwell on the mass mole quantities. ONE MOLE of all substances contains the same number of items. Turn this concept around a little bit. Given a mass quantity of a substance how can the number of moles it represents be determined? The number of moles in a given mass (weight in grams) is equal to that part of the stard mass it corresponds to. The stard mass of antimony (at.no. 51) is grams per mole. How many moles are contained in grams of antimony? (ANS: Compare the given mass of antimony (7.401 g) to the stard mass ( g). What part of the stard mass is the given mass? Or, how many times does the stard mass go into the given mass?, i.e., part / whole) Use the suggested method of setting up such conversions? moles Sb = g Sb 1 mole Sb g Sb = moles Sb (ARANF) SOLUTIONS are homogeneous mixtures that consist of three parts: (What is the third part?) one part is the solute (component present in lesser amount), another part is the solvent (component present in greater amount). In aqueous solutions the solvent is water. The solution process Ionic bonding exists only in the crystalline solid state. Consequently, when ionic substances are dissolved in water, ionic bonds are broken. This requires energy. The fact that aqueous solutions of ionic substances are quite common suggests that the solution process compensates for energy required to break ionic bonds. Part of this energy is provided by the free independent migration of oppositely charged ions in solution, which, in the solid, were locked in rigid positions. Additional energy is provided b/c of stabilizing interactions between ions water molecules. Water is a polar molecule, i.e., the molecule has a non-uniform distribution of charge, although there is no net charge. The more positive regions on water molecules (the H atoms) tend to orient towards anions in solution, the more negative regions (the O atoms) tend to orient towards cations in solution. This preferential orientation is the major driving force of the solution process for ionic solutes is represented in chemical formulas by appending a condition subscript of. So NaCl (s) refers to solid sodium chloride, whereas NaCl refers to an aqueous solution containing the separated ions of sodium chloride. (DEMO conductivity of electrolyte vs non-electrolyte aqueous solutions.) 1. Expressing CONCENTRATIONS of substances present in solutions The unit of concentration of interest at this time is the MOLARITY unit. Molarity is defined as a ratio of moles of solute over Volume of solution expressed in Liter units. The symbol for molarity is capital M, the math expression is... MOLARITY = (moles solute) / (vol.solution in liter units) M = moles / L A. Describe how to prepare: i. 500 ml of a 0.15 M solution of potassium iodide; ii L of M potassium dichromate. B. How many moles are in: i. 7.4 ml of a 0.17 M solution of sodium hypochlorite; ii. 6.7 x 10 - L of 0.08 M barium nitrate. C. How many ml of a M solution of HCl will contain: i x 10 - moles of hydrogen ion; ii grams of HCl. D. How many moles of ions are present in 4.9 ml of: i. 1.5 M barium hydroxide; ii M chromium(iii) chloride.

2 Notice that the product of (vol.solution in LITERS) x (conc. in MOLARITY) = moles solute, this is the path connecting solutions to all of the mass/number/mole relationships already met in Chapter. It will always be true that any two substances (say A B) in any reaction will be related as follows: (moles of A) = (moles of B) x (conversion factor(s)) This relation between moles A B provides the stoichiometric basis for every chemical reaction. (moles of A) = (moles of B) x (conversion factor(s)) Examples of Solution Stoichiometry Problems A. A M solution of nickel(ii) sulfate ( Ni SO 4 ) is reacted with a M solution of lead(ii) nitrate to form the insoluble substance lead(ii) sulfate. What volume of the lead(ii) nitrate solution is needed to react with ml of the nickel(ii) sulfate solution? B. Hydrochloric acid ( H Cl ) reacts with calcium hydroxide [ Ca ( OH ) ] in a "neutralization" reaction ( reaction of acid with base forming a "salt" water). In a one particular reaction 9.7 ml of the acid was needed to completely react with 5.00 ml of the base. If the concentration of the acid is known to be 0.15 M, what is the concentration of the base? C. A 0.56 M solution of silver(i) nitrate ( Ag NO ) is to be reacted with a M solution of scium(iii) chloride ( Sc Cl ) to form the insoluble substance silver(i) chloride ( Ag Cl ). What volume of the scium chloride solution will be needed to react completely with 4.15 ml of the silver(i) nitrate solution? D. In order to produce complete reaction, 5.00 ml of a lanthanum chloride solution ( La Cl ) are needed to react with 1.85 ml of a 0.05 M solution of sodium oxalate ( Na C O 4 ) to form the insoluble substance lanthanum oxalate [ La ( C O 4 ) ]. What is the molarity of the lanthanum chloride solution? What mass of precipitate would be formed?

3 Answers 1. Expressing CONCENTRATIONS of substances present in solutions Molarity = moles solute vol.solution in Liter units A. i. solve for moles of potassium iodide needed to form 500 ml of a 0.15 M solution. moles KI = ( molarity ) x (liters solutions) = (0.15 M ) x ( L ) = moles KI convert to grams KI: directions:? g KI = moles KI 166g( KI ) 1mole( KI ) = 1.45 g KI weigh out 1.45 g KI, add enough water to completely dissolve all solute, to attain a final combined volume of 500 ml. ii.? moles K Cr O 7 = (0.085 M) x (0.50 L) = moles K Cr O 7 convert to grams:? g K Cr O 7 = moles directions: 94g 1mole = g weigh out g K Cr O 7, add enough water to completely dissolve all solute, attain a final combined volume of 50 ml. B. i.? moles Na Cl O = ( 0.17 M ) x ( 0.74 L) = ii.? mole Ba ( N O ) = ( 0.08 M ) x (0.067 L) = C. i. hydrochloric acid is a STRONG acid, i.e., it is completely ionized in solution. Each HCl furnishes ONE hydrogen ion, so molarity HCl = molarity H 1+? L = ( 5.00 E - moles H 1+ ) / ( M ) = Liters or 5.71 ml ii. suggest converting 1.50 grams to moles HCl first, then finding volume of solution? moles HCl = 1.50 g HCl 1mol( HCl) 6. 5g( HCl) = ? L = ( moles HCl ) / ( M ) = L or ml D. i. Barium hydroxide, Ba ( OH ), is completely ionized in aqueous solutions, it is a STRONG base. The balanced equation showing this process would be: Ba ( OH ) (s) ---> Ba + + OH 1- Note that THREE ions are provided by each formula unit, so whatever the concentration of Ba ( OH ), the TOTAL concentration of ions will be THREE times as great. [ Ba ( OH ) ] = 1.5 M so [ ions ] =.75 M? moles ions = (.75 M in ions ) x ( L ) = 0.16 ii. Chromium(III) chloride, Cr Cl, is a salt will be completely ionized in aqueous solution. The balanced equation showing this process would be: Cr Cl (s) ---> Cr + + Cl 1- Note that FOUR ions are provided by each formula, so whatever the concentration of Cr Cl, the TOTAL concentration of ions will be FOUR times as great. [ Cr Cl ] = M so [ ions ] =.56 M? moles ions = (.56 M in ions ) x ( L ) = worked-out Examples of Solution Stoichiometry Problems follow:

4 A. Starting Information: M reactant NiSO 4 Ni + SO 50 ml M + ( aq) 4 ( aq) + 1 ( aq) ( aq) reactant Pb(NO ) Pb + NO? VOL Lead(II) cations sulfate anions + ( aq) 4 ( aq) 4 ( s) will form insoluble lead(ii) sulfate: Pb + SO PbSO moles Pb(NO ) = moles NiSO 4 [ C.F. ] ( M ) (? L) = ( M)(0.050 L)? Vol. in ml = mole SO4 1 mole NiSO 1 mole Pb 1 mole SO 1 mole Pb NO ( ) 1 mole Pb

5 B. Starting Information: 0.15 M reactant HCL H + Cl 9.7 ml? M reactant Ca( OH ) Ca + OH 5.00 ml Hydrogen cations hydroxide anions (liq.) will form water in a neutralization reaction: H + OH H O moles Ca( OH ) = moles HCl [ C.F. ] (? M ) (5.00 ml) = ( 0.15 M)(9.7 ml)? M = mole H 1 mole OH 1 mole HCl 1 mole H mole Ca(OH) 1- mole OH

6 C. Starting Information: 0.56 M reactant AgNO Ag + NO M 4.15 ml 1+ 1 ( aq) ( aq) + reactant ScCl Sc + Cl? VOL Silver(I) cations chloride anions 1 1+ will form insoluble silver(i) chloride: Ag + Cl AgCl (s) moles ScCl = moles ScCl [ C.F. ] ( M ) (? ml) = ( 0.56 M)(4.15 ml)? ml = mole Ag 1 mole AgNO 1 mole Cl 1 mole Ag mole ScCl 1 mole Cl

7 D. Starting Information:? M reactant LaCl La + Cl 5.00 ml 0.05 M reactant Na C O ml Lanthanum(III) cations oxalate anions + 1 Na 1+ ( aq) + CO 4 ( aq) + will form insoluble lanthanum(iii) oxalate: La + CO 4 La (C O ) 4 (s) moles LaCl = moles Na C O 4 [ C.F. ] (? M ) ( 5.00 m L) = ( 0.05 M)(1.85 ml)? M = mole CO4 1 mole Na C O mole La mole C O 1 mole LaCl 1 mole La

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