Ba(NO ) Ba + 2NO. NaF Na + F. Ba + 2F = BaF

Transcription

1 CHEM 105 COMBINED 101 & 105 DISC SECTIONS 4-OCT a. Consider.7 ml of a M solution of barium nitrate Write a net ionic equation depicting the aqueous solution of barium nitrate. 3 Ba(NO ) Ba + NO 3 the mole ratio of barium nitrate to barium cation is 1 : 1 b. Consider ml of a M sodium fluoride solution. Write a net ionic equation depicting the aqueous solution of sodium fluoride. NaF Na + F the mole ratio of sodium fluoride to fluoride anion is 1 : 1 c. When the two solution are mixed a precipitate of barium fluoride forms. Write the net ionic equation for the precipitation reaction. Ba + F = BaF d. Calculate the millimoles of each reactant (the reactant ions are Ba and F ) mmoles Ba(NO 3 ) = mmoles Ba = ( M ) ( ml ) = ( M )(.7 ml ) =.6584 mmoles NaF = mmoles F = ( M ) ( ml ) = (s) e. Set-up a B4, REACT and AFTR grid for this reaction and fill-in all cells Ba (limiting) F (XS) BaF B none REACT (-1 : - : +1) AFTR none f. What mass of precipitate will be formed? mmoles = mass (mg) / std.mass? mg = mmoles x std.mass = X (137+x19) = 465 mg = grams g. What is the concentration of reactant ion that is present in an excess amount, after the reaction has completed? Molarity = mmoles / ml soln. = / ( ) = molar in F

2 . a. Consider a M sodium hydroxide. Write a net ionic equation depicting an aqueous solution of sodium hydroxide. This is a strong base and the only species present in solution are the ions. NaOH Na + OH b. Consider an oxalic acid solution of unknown molarity. Write a net ionic equation depicting an aqueous solution of oxalic acid. Oxalic acid is a a dibasic acid, H C O 4. This is an organic acid, i.e., a weak acid. The major species present in solution is molecular unionized oxalic acid. 4 H C O H + C O 4 c. Write a net ionic equation showing the reaction of these two solutions. The acid is dibasic so TWO hydroxide anions are required for each oxalic acid molecule. 4 4 H C O + OH = C O + H O d. Titration of 5.00 ml of the base solution requires 7.19 ml of the acid solution. What is the molarity of the oxalic acid solution? mmoles oxalic acid = mmoles NaOH 1 1 mole OH 1 mole oxalic acid 1 1 mole NaOH mole OH The first conversion factor comes from part.a., and the second from.c. (?M)(7.19mL)=(0.147M)(5.00mL)[ 1/1 ][ 1/ ]?M = M oxalic acid 3. a. Consider.051 grams of lead nitrate. It is dissolved in enough water to form 50 ml of solution. Write a net ionic equation depicting an aqueous solution of 3 lead nitrate. Pb(NO ) Pb + NO M (lead nitrate) =.051 g / ( 331 g / mole) 0.50 L soln 3 = M b. Consider grams of barium chloride. It is dissolved in enough water to form 00 ml of solution. Write a net ionic equation depicting an aqueous solution of barium chloride. BaCl Ba + Cl M (barium nitrate) = g / ( 08 g / mole) 0.00 L soln = M c. When these two solutions are mixed, a precipitate of lead chloride forms. Write a net ionic equation showing this reaction. Pb + Cl PbCl d. If ml of the barium chloride solution are titrated against the lead nitrate solution, what volume of the latter would be required? mmoles Pb(NO 3 ) = mmoles BaCl 1 mole Cl 1mole BaCl (s) 1mole Pb NO ( ) 1mole Pb 1mole Pb mole Cl? ml = {(14.85 ml x M ) / ( M)} [ / 1 ][ 1 / ][ 1 / 1 ]? ml = 13.58

3 4. a. Consider.78 ml of a M solution of aniline, C 6 H 5 NH. Write a net ionic equation depicting an aqueous solution of aniline. Aniline is a weak base. The major specie present in solution is molecular unionized aniline C H NH + HOH C H NH + OH b. Consider ml of a M solution of sulfuric acid, H SO 4. Write a net ionic equation depicting an aqueous solution of sulfuric acid. This is a dibasic acid, and the ionization occurs in a stepwise fashion. The first ionization is complete, but the second is not. However, in the forthcoming neutralization reaction it can be assumed that sufficient base is present to react with BOTH hydrogen ions of sulfuric acid. So, write the net ionic equation to show both hydrogen as being ionized. 4 H SO H + SO c. Write a net ionic equation for the reaction of these two solutions C H NH + H C H NH 4 d. determine the millimole quantities of each reactant in the above equation. mmoles aniline = (0.085M)(.78mL) = mmoles H SO 4 = (0.107M)(16.88mL) = , so: mmoles H = e. Set-up a B4, REACT and AFTR grid for this reaction and fill-in all cells. C 6 H 5 NH (lim) H (XS) C 6 H 5 NH 3 B none REACT (-1:-1:+1) AFTR none g. What is the molarity of the reactant present in an excess amount after the reaction is complete? [ H ] = molarity hydrogen ion = mmoles / ml = ( ) / ( ) [ H ] = M

4 5. Consider the skeletal equation below: H O + MnO = O + Mn (acid) 4 (g) a. Construct a balanced equation for the oxidation half-cell. H O = O + e + H (g) 1 b. Construct a balanced equation for the reduction half-cell. (Aq) H + MnO + 5 e = Mn + 4 H O c. Construct a balanced equation for the overall reaction. 6. Consider the skeletal equation below FeO = Fe O + HO (s) (base) a. Construct a balanced equation for the reduction half-cell (s) FeO + 6 e + 5 H O = Fe O + 10 OH b. Construct a balanced equation for the oxidation half-cell. What is present in this solution that could react to form peroxide anion? Water? (yes) Hydroxide anion? (yes) Oxygen? (NO!) so start with OH = HO + e 1 then add another hydroxide to balance charge. 1 3 OH = HO + e and finally add water to balance mass 1 3 OH = HO + e + H O c. Construct a balanced equation for the overall reaction.

5 7. W / R to the reaction whose equation was balanced in problem 5: Suppose that ml of a 30% solution of hydrogen peroxide (density = 1.14 g / ml) was reacted with L of a 0.11 M potassium permanganate solution. a. determine the millimoles of each reactant. H O ml of solution have a mass of: ml x 1.14 g / ml = 34.0 grams SOLUTION 30% of this mass is H O, or 0.30 x 34.0 = 10.6 grams H O millimoles H O = mass (mg) / Std.Mass = 1060 / 34 = millimoles KMnO 4 = millimoles MnO 4 = (0.11 M)(1000mL) = 11 b. Set-up a B4, REACT and AFTR grid and fill-in all cells. The balanced redox equation informs that 5 moles hydrogen peroxide reacts with moles permanganate anion to form 5 moles molecular oxygen. Work a "trial" calculation:? mmoles MnO 4 = H O mmoles MnO 5 mmoles H O 4 = 10.7 So, 10.7 mmoles MnO 4 will be needed to react with all (301.76) of the H O present. But, only 11 mmoles MnO 4 are present. So, MnO 4 is the limiting reagent and H O is present in an excess amount. Work another "trial" calculation, this time to find how much H O will be used to react with all of the limiting reagent, i.e., the 11 mmoles of MnO 4. 5 mmoles H O? mmoles H O = 11 mmoles MnO 4 mmoles MnO 4 = 80 Enter this quantity in the proper cell and then fill-in all remaining cells. 5 H O (XS) MnO 4 (lim) 5 O B none REACT (-5:-:+5) AFTR 1.76 none 80 c. What mass of oxygen gas would be formed? moles = mass / Std.Mass so mass = moles x Std.Mass? mg oxygen = 80 mmoles oxygen x ( 3 mg / mmole ) = 8,960 mg

6 or (convert to grams): grams oxygen gas d. What is the molarity of the reactant present in an excess amount after the reaction is complete? Molarity = moles / L soln = mmoles / ml soln [ H O ] = 1.76 / ( ) = M 8. W/ R to the reaction whose equation was balanced in problem 6: Suppose that ml of a M solution of sodium ferrate, Na FeO 4, undergoes this reaction. a. What mass of iron(iii) oxide would be formed? The balanced chemical equation shows that TWO moles ferrate anion (or TWO moles of the compound sodium ferrate - because the anion and compound are in a 1 : 1 mole ratio) form ONE mole iron(iii) oxide. This is the conversion factor that relates these substances in this reaction. moles iron(iii) oxide = moles sodium ferrate [ C.F. ]? g FeO3 1 mole FeO3 = ( M)( L) Std. Mass mole Na FeO 4? g Fe O 3 = 0.07 grams b. If the solution was neutral before the reaction, what would be the concentration of hydroxide anion after reaction? The balanced chemical equation shows that ONE mole of hydroxide anions are formed from TWO moles of sodium ferrate. mmoles sodium ferrate = ( M ) ( ml ) = ( M ) ( ml ) = mmoles hydroxide anion = mmoles sodium ferrate [ 1 / ] = [ OH 1 - ] = molarity hydroxide anion = mmoles / ml = / ml [ OH 1 - ] = molar