Solutions for Practice Problems. Section 9.1 Making Predictions About Solubility. Ions. Formula. PbCl 2. Pb 2+ Cl. Zn 2+ O 2. ZnO.

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1 Chapter 9 Aqueous Solutions Solutions for Practice Problems Section 9.1 Making Predictions About Solubility Student Textbook page Problem Decide whether each of the following salts is or in distilled water. Give reasons for your answer. (a) lead(ii) chloride, PbCl (a white crystalline powder used in paints) (b) zinc oxide, ZnO (a white pigment used in paints, cosmetics, and calamine lotion) (c) silver acetate, AgCH 3 COO (a whitish powder that is used to help people quit smoking because of the bitter taste it produces) Determine if the salts listed will dissolve in water. You have the names of the three salts and access to the solubility guidelines. Identify the cations and anions in each salt. Refer to the solubility chart on page 334 of the text and determine the guideline solubility of these ions. The solubility of the salt will correspond with the solubility of the ion having the higher guideline (lower number). Name Formula Ions Guideline Ion Solubility Salt Solubility lead(ii) chloride PbCl Pb Cl 3 INSOLUBLE zinc oxide ZnO Zn O 5 INSOLUBLE silver acetate AgCH 3 COO Ag CH 3 COO 1 SOLUBLE Check your results against another reference. These conclusions are correct.. Problem Which of the following compounds are in water? Explain your reasoning for each compound. (a) potassium nitrate, K (used to manufacture gunpowder) (b) lithium carbonate, Li CO 3 (used to treat people who suffer from depression) (c) lead(ii) oxide, PbO (used to make crystal glass) Determine if the salts listed will dissolve in water. 154

2 You have the names of the three salts and access to the solubility guidelines. Identify the cations and anions in each salt. Refer to the solubility chart on page 334 of the text and determine the guideline solubility of these ions. The solubility of the salt will correspond with the solubility of the ion having the higher guideline (lower number). Name Formula Ions Guideline Ion Solubility Salt Solubility potassium nitrate K K 1 1 SOLUBLE Lithium carbonate Li CO 3 Li CO 3 1 SOLUBLE lead(ii) oxide PbO Pb O INSOLUBLE Check the predictions against another reference. These results are correct. 3. Problem Which of the following compounds are in water? (a) calcium carbonate (present in marble and limestone) (b) magnesium sulfate, Mg (found in the hydrated salt, Mg 7H O, also known as Epsom salts; used for the relief of aching muscles and as a laxative) (c) aluminum phosphate, AlPO 4 (found in dental cements) Determine if the salts listed will dissolve in water. You have the names of the three salts and access to the solubility guidelines. Identify the cations and anions in each salt. Refer to the solubility chart on page 334 of the text and determine the guideline solubility of these ions. The solubility of the salt will correspond with the solubility of the ion having the higher guideline (lower number). Name Formula Ions Guideline Ion Solubility Salt Solubility calcium carbonate CaCO 3 Ca CO 3 4 INSOLUBLE magnesium sulfate Mg Mg 5 5 SOLUBLE aluminum phosphate AlPO 4 Al 3 3 PO 4 5 INSOLUBLE Check the predictions against another reference. These results are correct. 155

3 Section 9. Reactions in Aqueous Solutions Student Textbook page Problem Predict the result of mixing each pair of aqueous solutions. Write a balanced chemical equation if you predict that a precipitate forms. Write NR if you predict that no reaction takes place. (a) sodium sulfide and iron(ii) sulfate (b) sodium hydroxide and barium nitrate (c) cesium phosphate and calcium bromide (d) sodium carbonate and sulfuric acid (e) sodium nitrate and copper(ii) sulfate (f) ammonium iodide and silver nitrate (g) potassium carbonate and iron(ii) nitrate (h) aluminum nitrate and sodium phosphate (i) potassium chloride and iron(ii) nitrate (j) ammonium sulfate and barium chloride (k) sodium sulfide and nickel(ii) sulfate (l) lead(ii) nitrate and potassium bromide Predict whether or not each pair of aqueous solutions will provide ions that will combine to form an product (precipitate). Write a balanced chemical equation if a reaction is predicted and NR if no reaction is predicted. You know the names of the compound in each solution. Identify the ions in each compound. Exchange the cations in the two compounds and for each new compound, look up its solubility guideline on page 334 of the textbook. Predict whether or not either of these new compounds is. Write the balanced equation for those examples in which a new product formed and NR if both the new compounds are. Solubility Key: S = I = (a) sodium sulfide iron(ii) nitrate Reactant Ions Na S Fe Products Ions Na Fe S Guideline Ion Solubility S S S I Product Na FeS Product Solubility S I Na S (aq) Fe(aq) Na (aq) FeS (s) (b) sodium hydroxide barium nitrate Reactant Ions Na OH Ba Products Ions Na Ba OH Guideline Ion Solubility S S I I Product Na Ba(OH) Product Solubility S S (exception to rule) 156

4 NaOH (aq) Ba( ) (aq) NR (c) cesium phosphate calcium bromide Reactant Ions Cs 3 PO 4 Ca Br Products Ions Cs Br Ca 3 PO 4 Guideline Ion Solubility S S I I Product CsBr Ca 3 PO 4 Product Solubility S I Cs 3 PO 4(aq) 3CaBr (aq) 6CsBr (aq) Ca 3 (PO 4 ) (s) (d) sodium carbonate and sulfuric acid Reactant Ions Na CO 3 H Products Ions Na H CO 3 Guideline Ion Solubility S S S I Product Na H CO 3 Product Solubility S S Na CO 3(aq) H (aq) NR (e) sodium nitrate and copper(ii) sulfate Reactant Ions Na Cu Products Ions Na Cu Guideline Ion Solubility S S S S Product Na Cu( ) Product Solubility S S Na(aq) Cu(aq) NR (f) ammonium iodide and silver nitrate Reactant Ions NH 4 I Ag Products Ions NH 4 Ag I Guideline Ion Solubility S S I S Product NH 4 AgI Product Solubility S S NH 4 I (aq) Ag(aq) NH 4 (aq) AgI (s) (g) potassium carbonate and iron(ii) nitrate Reactant Ions K CO 3 Fe Products Ions K Fe CO 3 Guideline Ion Solubility S S S I Product K FeCO 3 Product Solubility S I K CO 3(aq) Fe( ) (aq) K(aq) FeCO 3(s) (h) aluminum nitrate and sodium phosphate Reactant Ions Al 3 Na 3 PO 4 Products Ions Al 3 3 PO 4 Na Guideline

5 Ion Solubility S I S S Product AlPO 4 Na Product Solubility I I Al( ) 3(aq) Na 3 PO 4(aq) 3Na(aq) AlPO 4(s) (i) potassium chloride and iron(ii) Reactant Ions K Cl Fe Products Ions K Fe Cl Guideline Ion Solubility S S S S Product K FeCl Product Solubility S S KCl (aq) Fe( ) (aq) NR (j) ammonium sulfate and barium chloride Reactant Ions NH 4 Ba Cl Products Ions NH 4 Cl Ba Guideline Ion Solubility S S I S Product NH 4 Cl Ba Product Solubility S I (NH 4 ) (aq) BaCl (aq) NH 4 Cl (aq) Ba(s) (k) sodium sulfide and nickel(ii) sulfate Reactant Ions Na S Ni Products Ions Na Ni S Guideline Ion Solubility S S S I Product Na NiS Product Solubility S I Na S (aq) Ni(aq) Na (aq) NiS (s) (l) lead(ii) nitrate and potassium bromide Reactant Ions Pb K Br Products Ions Pb Br K Guideline Ion Solubility I S S S Product PbBr K Product Solubility I S Pb( ) (aq) KBr (aq) K(aq) PbBr (s) Examine the final equation to see if the ionic compounds that are noted as being in aqueous solution are and the compounds noted as solid are. Solutions for Practice Problems Student Textbook page Problem Mixing each pair of aqueous solutions results in a chemical reaction. Identify the spectator ions. Then write the balanced net ionic equation. 158

6 (a) sodium carbonate and hydrochloric acid (b) sulfuric acid and sodium hydroxide Identify the spectator ions and write a balanced net ionic equation. You know the chemical names of the compounds. Write the chemical formula of each compound, then complete the chemical equation for the reaction. Using the solubility guideline, note which product is and which is the precipitate. Replace the formulae of the ionic compounds with dissociated ions. Identify the ions that appear on both sides of the equation as spectator ions. Rewrite the equation without the spectator ions. (a) Na CO 3(aq) HCl (aq) NaCl (aq) H CO 3(aq) H CO 3(aq) CO (g) H O (l) Na (aq) CO 3 (aq) H (aq) Cl (aq) Na (aq) Cl (aq) CO (g) H O (l) spectator ions are: Na (aq) and Cl (aq) net ionic equation: H (aq) CO 3 (aq) CO (g) H O (l) (b) H (aq) NaOH (aq) Na (aq) H O (l) H (aq) (aq) Na (aq) OH (aq) Na (aq) (aq) H O (l) spectator ions are: Na (aq) and (aq) net ionic equation: H (aq) OH (aq) H O (l) Examine the net ionic equation to confirm that no ions are common to both sides and that the equation is balanced. 6. Problem Identify the spectator ions for the reaction that takes place when each pair of aqueous solutions is mixed. Then write the balanced net ionic equation. (a) ammonium phosphate and zinc sulfate (b) lithium carbonate and nitric acid (c) sulfuric acid and barium hydroxide Identify the spectator ions and write a balanced net ionic equation. You know the chemical names of the compounds. Write the chemical formula of each compound, then complete the chemical equation for the reaction. Using the solubility guideline, note which product is and which is the precipitate. Replace the formulae of the ionic compounds with dissociated ions. Identify the ions that appear on both sides of the equation as spectator ions. Rewrite the equation without the spectator ions. (a) (NH 4 ) 3 PO 4(aq) 3Zn(aq) Zn 3 (PO 4 ) (s) 3(NH 4 ) (aq) 3 6NH 4 (aq) PO 4 (aq) 3Zn (aq) 3 (aq) 3Zn 3 (aq) PO 4 (aq) 6NH 4 (aq) 3 (aq) spectator ions are: NH 4 (aq) and (aq) net ionic equation: 3Zn 3 (aq) PO 4 (aq) Zn 3 (PO 4 ) (s) (b) Li CO 3(aq) H(aq) Li(aq) H O (l) CO (g) 159

7 Li (aq) CO 3 (aq) H (aq) (aq) Li (aq) (aq) H O (l) CO (g) spectator ions are: Li (aq) and (aq) net ionic equation: H (aq) CO 3 (aq) CO (g) H O (l) (c) H (aq) Ba(OH) (aq) Ba(s) H O (l) H (aq) (aq) Ba (aq) OH (aq) Ba(s) H O (l) spectator ions are: none net ionic equation: H (aq) (aq) Ba (aq) OH (aq) Ba(s) H O (l) Examine the net ionic equation to confirm that no ions are common to both sides and that the equation is balanced. Section 9.3 Stoichiometry in Solution Chemistry Student Textbook page Problem Food manufacturers sometimes add calcium acetate to puddings and sweet sauces as a thickening agent. What volume of mol/l calcium acetate, Ca(CH 3 COO) (aq), contains mol of acetate ions? Find the volume of mol/l calcium acetate, Ca(CH 3 COO) (aq), solution that contains mol of acetate ions? You know the concentration of the calcium acetate solution, the formula for calcium acetate, and the required number of moles of acetate ion. Write the equation for the dissociation of Ca(CH 3 COO). Determine the ratio of moles CH 3 COO (aq) : Ca(CH 3COO) to calculate the moles of Ca(CH 3 COO) (aq). Rearrange the formula n = C V and calculate the volume of solution required. Ca(CH 3 COO) Ca (aq) CH 3 COO (aq) CH 3 COO 1 mol Ca(CH 3COO) = mol Ca(CH mol CH 3 COO 3 COO) V = n C = mol mol/l = L or 300 ml of Ca(CH 3COO) (aq) The final answer has the correct unit and number of significant figures. The answer seems to be reasonable. 8. Problem Ammonium phosphate can be used as a fertilizer. 6.0 g of ammonium phosphate is dissolved in sufficient water to produce 300 ml of solution. What are the concentrations (in mol/l) of the ammonium ions and phosphate ions present? 3 Find the concentrations of the NH 4 (aq) and PO 4 (aq) ions in solution. You know the mass of ammonium phosphate and the final volume of the solution. 160

8 Calculate the molar mass of (NH 4 ) 3 PO 4 and convert 6.0 g of this compound to moles. Calculate the concentration of (NH 4 ) 3 PO 4 in mol/l.write the balanced equation for the dissociation of (NH 4 ) 3 PO 4 and use the mole ratios in this equation to determine the concentration of each of the ions. molar mass of (NH 4 ) 3 PO 4 = g/mol 6.0 g (NH 4 ) 3 PO 4 1 mol g = mol (NH 4) 3 PO mol concentration of (NH 4 ) 3 PO 4 = = 0.13 mol/l L 3 (NH 4 ) 3 PO 4(s) 3NH 4 (aq) PO 4 (aq) moles NH 4 (aq) = 0.13 mol/l (NH 4 ) 3 PO 4 = 0.39 mol NH 4 (aq) moles PO 4 3 (aq) = 0.13 mol (NH 4 ) 3 PO 4 = 0.13 mol PO 4 3 (aq) seems reasonable. 9. Problem An aqueous solution of a certain salt contains chloride ions. A sample of this solution was made by dissolving g of the salt in a 1 L volumetric flask. Ten 5.00 ml of the solution was treated with excess silver nitrate. The precipitate, AgCl, was filtered and dried. If the mass of this dry precipitate was g, what was the mass percent of chloride ions in the solution? Determine the mass percent of chloride ion in a salt solution. You know the original mass of salt used to prepare the solution. Also, you know the mass of AgCl that precipitates from a 5.00 ml sample of the solution reacts with an excess of Ag. Determine the molar mass of AgCl and convert g of this compound to moles. From the mol ratio of Cl in AgCl, determine the moles of chloride ion in the precipitate. All of the chloride from the 5.00 ml sample of salt solution is in this AgCl. Calculate the moles of Cl in L of solution and use the molar mass of Cl to convert this to grams. Use this mass and the original mass of the salt solution to determine the mass percent of Cl in the solution. molar mass AgCl = g/mol mass of Cl in precipitate = mass of Cl in 5.00 ml sample 1 mol = g AgCl = g Cl mass of Cl per L = mol Cl mass percent of Cl in original solution = g 1000 ml 5.00 ml g Cl 1 mol mass of Cl mass of sample 100% = 4.68 g g 100% = 6.6% = 4.68 g seems reasonable. 161

9 10. Problem The active ingredient in some rat poisons is thallium(i) sulfate, Tl. A chemist takes a 500 mg sample of thallium(i) sulfate and adds potassium iodide, to precipitate yellow thallium(i) iodide. When the precipitate is dried, its mass is 00 mg. What is the mass percent of Tl in the rat poison? Calculate the mass percent of Tl in 500 mg sample of rat poison. You know the original mass of the rat poison, the mass of a precipitate of TlI and the chemical formula of thallium sulfate. All of the thallium in the precipitate TlI was in the original sample of rat poison. Determine the molar mass of TlI and calculate the moles of Tl in the 00 mg of the precipitate. Write the balance equation for the reaction between Tl and KI and use the mole ratio in this equation to calculate the moles of Tl. Find the molar mass of Tl and calculate the mass of this compound. Use this mass and the mass of the sample of rat poison to determine the mass percent of Tl in the rat poison. molar mass of TlI = g/mol Tl (aq) KI (aq) TlI (s) K (aq) moles Tl = 0.00 g TlI 1 mol TlI g TlI 1 mol Tl mol TlI = mol Tl mol Tl g = 0.15 g Tl 1 mol mass percent Tl in sample = mass of Tl mass of sample 100% = 0.15 g g 100% = 30.5% seems reasonable based upon the mass of the sample. Solutions for Practice Problems Student Textbook page Problem 8.76 g of sodium sulfide is added to 350 ml of 0.50 mol/l lead(ii) nitrate. Calculate the maximum mass of precipitate that can form. You must calculate the mass of precipitate that forms when known amounts of solutions Na S and Pb( ) react. You know a mass of Na S and a volume and concentration of Pb( ) solution. Find the molar mass of Na S and calculate the moles of this compound. Use the formula n = C V to calculate the moles of Pb( ). Write the balanced equation for the reaction between these two compounds and refer to the solubility guideline to identify the product. Identify which of these two reactants is the limiting 16

10 reagent. Use the limiting reagent to calculate the moles of precipitate. Determine the molar mass of the precipitate and convert the moles to grams. molar mass of Na S = g/mol n = = = 0.11 mol Na S mass molar mass 8.76 g g/mol n = C V = 0.50 mol/l L = mol Pb( ) Na S (aq) Pb( ) (aq) Na(aq) PbS (s) The precipitate is PbS. Since the reactants react in a molar ratio of 1:1, the compound with the lesser number of moles will be the limiting reagent, i.e. the mol Pb( ). Also, since the mol ratio of Pb( ) : PbS is 1:1, we can conclude that mol of Pb( ) will produce mol of PbS. molar mass of PbS = 39. g/mol mass of PbS mol 39. g = 0.9 g of PbS 1 mol seems reasonable based upon the mass of the sample. 1. Problem 5.0 ml of mol/l Pb( ) (aq) is mixed with 300 ml of 0.0 mol/l KI (aq). What is the maximum mass of precipitate that can form? Find the mass of precipitate that will form when known volumes of two solutions of given concentration are mixed? You are given the volumes, concentrations, names and formulae of two solutions that are to be mixed. Write the balanced equation for this reaction and identify the product. Use the formula n = C V to calculate the number of moles of each reactant. Identify the limiting reagent. Use the limiting reagent to calculate the moles of precipitate. Find the molar mass of the precipitate and convert the moles to grams. Pb( ) (aq) KI (aq) K(aq) PbI (s) moles Pb( ) (aq) = n = C V = mol/l L = mol moles KI (aq) = n = C V = 0.0 mol/l L = mol From the balanced equation, Pb( ) : KI = 1 : mol Pb( ) mol KI 1 mol Pb( ) = mol KI But there are mol of KI available, therefore KI is in excess and Pb( ) is the limiting reagent. From the balanced equation, the molar ratio Pb( ) : PbI is 1:1. Therefore, mol of PbI are produced. molar mass of PbI = g/mol mass of PbI = mol g mol = 4.61 g PbI seems reasonable based upon the mass of the sample. 163

11 13. Problem A student mixes 15.0 ml of 0.50 mol/l aqueous sodium hydroxide with 0.0 ml of mol/l aqueous aluminum nitrate. (a) Write the chemical equation for the reaction. (b) Calculate the maximum mass of precipitate that forms. For the reaction between two solutions, write the balanced equation and determine the mass of precipitate that will form. You are given the volumes, concentrations, names, and chemical formulae of the two solutions. (a) This reaction is a double displacement reaction that follows the form: AB CD CB AD. Refer to Table 9.1 General Solubility Guidelines and determine which product is a precipitate. Write the chemical formula for each compound and balance the equation. (b) Rewrite the balanced equation showing each ionic compound dissociated into its ions. Write the net ionic equation for the reaction. Calculate the number of moles of each reactant ion. Identify the limiting reagent. Use the limiting reagent to determine the number of moles of precipitate that form. Calculate the molar mass of the precipitate and convert the number of moles to grams. (a) 3NaOH (aq) Al( ) 3(aq) 3Na(aq) Al(OH) 3(s) (b) 3Na (aq) 3OH (aq) Al 3 (aq) 3 (aq) 3Na (aq) 3 (aq) Al(OH) 3(s) net ionic equation: Al 3 (aq) 3OH (aq) Al(OH) 3(s) Al(OH) 3 is the precipitate. moles Al 3 1 mol Al = mol/l Al(NO 1 mol Al( ) 3 ) L = mol 3 moles OH 1 mol OH = 0.50 mol/l L = mol 1 mol NaOH Since Al 3 : OH is 1:3, OH will produce less Al(OH) 3 and is the limiting reagent. moles of Al(OH) 3 = 1 mol Al(OH) mol OH = mol 3 mol OH mass of Al(OH) 3 = mol g = g 1 mol seems reasonable based upon the mass of the sample. 164

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