Chapter 9 Practice Worksheet: Reactions in Aqueous Solutions 1. The compound H 2 S is classified as a weak electrolyte. Describe/draw how it reacts when placed in water. Completely dissociates in water. When dissolves in water, the ions separate from each other to form a solution that conducts an electrical current. 2. The compound HNO 3 is classified as a strong electrolyte. Describe/draw how it reacts when placed in water. Only partially dissociates in water. Most molecules will stay bonded together but some will dissociate to produce ions (surrounded by water) 3. The compound C 11 H 22 O 11 is classified as a non electrolyte. Describe/draw how it reacts when placed in water. Does not dissociate in water. Will stay bonded together as a complete molecule when placed in water. When dissolves in water, molecules become separated from each other, but the molecules themselves remain intact. 4. Determine if the following compounds will be soluble or insoluble in water: CaSO 4 Insoluble K 2 O FeCl 2 Ag 2 PO 4 Insoluble Mg(ClO 4 ) 2 Na 3 PO 3 For the following double-displacement reactions, complete the equation and determine if there are any insoluble products (precipitates). If there is a precipitate, write the balanced ionic and net ionic equations. If there is no precipitate, write the balanced molecular and ionic equations. 5. LiBr (aq) + KCl (aq) LiCl (aq) + KBr (aq) Li + (aq) + Br - (aq) + K + (aq) + Cl - (aq) Li + (aq) + Cl - (aq) + K + (aq) + Br - (aq) No Net Ionic Chapter 9 Worksheet page 1 of 5
6. _2_ NaOH (aq) + Fe(C 2 H 3 O 2 ) 2 (aq) Fe(OH) 2 (s) + 2NaC 2 H 3 O 2 (aq) 2Na + (aq) + 2OH - (aq) + Fe 2+ (aq) + 2C 2 H 3 O 2 - (aq) 2Na + (aq) + 2 C 2 H 3 O 2 - (aq) + Fe(OH) 2 (s) Fe 2+ (aq) + 2OH - (aq) Fe(OH) 2 (s) 7. _3_ CrBr 2 (aq) + _2_ (NH 4 ) 3 PO 4 (aq) Cr 3 (PO 4 ) 2 (s) + 6 NH 4 Br(aq) 3Cr 2+ (aq) + 6Br - (aq) + 6NH 4 + (aq) + 2PO 4 3- (aq) 6NH 4 2+ (aq) + 6Br - (aq) + Cr 3 (PO 4 ) 2 (s) 3Cr 2+ (aq) + 2PO 4 3- (aq) Cr 3 (PO 4 ) 2 (s) 8. _3_ Mg(NO 3 ) 2 (aq) + _2_ Na 3 PO 4 (aq) 6 NaNO 3 (aq) + Mg 3 (PO 4 ) 2 (s) 3Mg 2+ (aq) + 6NO 3 - (aq) + 6Na + (aq) + 2PO 4 3- (aq) 6Na + (aq) + 6NO 3 - (aq) + Mg 3 (PO 4 ) 2 (s) 3Mg 2+ (aq) + 2PO 4 3- (aq) Mg 3 (PO 4 ) 2 (s) 9. Identify each of the following substances as acids or bases (or both): H 2 S Acid NaOH Base HCl Acid H 3 PO 3 Acid Ba(OH) 2 Base H 2 O Both (Amphoteric) NH 3 Base 10. Identify the oxidation numbers of each element in the following compounds or ions: H 2 SO 4 H: +1 S: +6 O: -2 H 2 O 2 H: +1 O: -1 Br 2 (l) Br: 0 Cd (s) Cd: 0 BaCO 3 Ba: +2 C: +4 O: -2 FeSO 4 Fe: +2 S: +6 O: -2 CuCl 2 Cu: +2 Cl: -2 NO 2 - N: +3 O: -2 NiNO 3 Ni: +1 N: +5 O: -2 Sr(NO 3 ) 2 Sr: -2 N: +5 O: -2 11. Consider the 5 main types of chemical reactions: synthesis, decomposition, combustion, single replacement and double replacement. Which one(s) are classified also as redox reactions and how can you tell? Combination, Synthesis, Decomposition, and Combustion. Neutralization and Double Replacement cannot be redox because there are no elements oxidized or reduced. Chapter 9 Worksheet page 2 of 5
12. For the following redox reactions, identify the species being oxidized, the species being reduced, the oxidizing agent, and the reducing agent: Fe (s) + F 2 (g) FeF 3 (s) Fe is oxidized/reducing agent F 2 is reduced/oxidizing agent Cu(ClO 3 ) 3 (aq) + Mg (s) Cu (s) + Mg(ClO 3 ) 2 (aq) Mg is oxidized/reducing agent Cu 3+ in Cu(ClO 3 ) 3 is reduced/cu(clo 3 ) 3 is oxidizing agent Li (s) + H 2 O (l) LiOH (aq) + H 2 (g) Li is oxidized/reducing agent H + in H 2 O is reduced/h 2 O is oxidizing agent 16. When do you use the Solubility Rules and when do you use the Activity Series of Metals? Use solubility rules to predict precipitates in double replacement reactions. Use Activity Series in cationic single replacement reactions. 17. Suppose you wanted to make a 0.250 M solution of KMnO4 using a 250 ml volumetric flask. Explain in procedural form how you would do this. Using the molarity formula (M = mol/l) substitute 0.250 M and 0.250 L into the equation to solve for number of moles of solute (0.0625 moles of KMnO 4 ). Then determine how much mass this is (0.0625 moles * 158.04 g/mol = 9.88 g KMnO 4 ). Then you would mass this out on a balance and pour it into the volumetric flask. Add about ¼ full of DI water and swish/invert around to allow solute to begin dissolving. Add more DI water so flask is about ½ filled, and mix thoroughly again. Finally, fill up to 250 ml mark on flask and invert flask several times to ensure proper mixing. 18. You need to make 250. ml of a 0.450 M solution of Sodium Hydroxide (NaOH). What mass of NaOH do you need to use? (0.450 M)(0.250 L) = 0.1125 moles NaOH (0.1125 moles NaOH)(40.00 g/mol) = 4.50 g Chapter 9 Worksheet page 3 of 5
19. Suppose you are given a stock bottle of concentrated HCl and are asked to prepare a 500. ml solution that is 0.100 M HCl. If the reagent bottle is labeled 12.0 M HCl, how much HCl do you use? How much water did you add? M 1 V 1 = M 2 V 2 (12.0 M)(V 1 ) = (0.100 M)(500. ml) V 1 = 4.17 ml of HCl If the total solution after dilution is 500. ml and 4.17 ml of that solution is HCl, then 500. 4.17 ml = 496 ml of H 2 O 20. Suppose you are given 0.800 L of a 0.750 M solution of K 2 SO 4. How many ions of K + are there in the solution? If [K 2 SO 4 ] = 0.750 M, then [K + ] = 1.50 M due to the fact that 1 mole of K 2 SO 4 produces 2 moles of K + ions. So (1.50M)*(0.800 L) = 0.1875 moles of K + (0.1875 moles K + )*(6.022 x 10 23 ions K + ) = 1.13 x 10 23 ions of K + 21. What volume of 0.398 M sulfuric acid is needed to neutralize 27.0 ml of a 0.419 M solution of potassium hydroxide? Sulfuric acid = H 2 SO 4 Potassium hydroxide = KOH S a M a V a = S b M b V b (2)(0.398 M)(V a ) = (1)(0.419 M)(27.0 ml) V a = 14.2 ml -OR- H 2 SO 4 + 2KOH K 2 SO 4 + 2H 2 O (27.0 ml KOH)*(1 L / 1000 ml)*(0.419 mol KOH / 1 L)*(1 mol H 2SO 4 / 2 mol KOH)*(1 L / 0.398 mol H 2SO 4)*(1000 ml / 1 L) = 14.2 ml 22. What is the molar mass of a diprotic acid if 28.5 ml of 0.1672 M NaOH is required to neutralize a 0.0812-g sample? If the acid is binary, what is the identity of the acid? We are ultimately solving for molar mass, which is units of (g/mol). We already know the grams we re dealing with (0.0812 grams of acid), so all we need to figure out is how many moles of acid this is. A diprotic acid means it has two ionizable hydrogens at the front of its formula. Since we are dealing with a base that has only one hydroxide, that means we are going to need twice as many moles of base as we need acid (in other words, we need half the amount of moles of acid as we do base). (0.1672 M)(0.0285 L) = 0.0047652 moles of NaOH / 2 = 0.0023826 moles of Acid Chapter 9 Worksheet page 4 of 5
Now we can take our grams (0.0812 g) and divide it by the moles (0.0023826 mol) to get our molar mass: 0.0812 g / 0.0023826 mol = 34.1 g/mol Now to find the identity of the acid, we are told it s a binary acid, meaning it s H 2 bonded to some single element. If the acid s overall molar mass is 34.1 and we know the molar mass of the two hydrogens is approximately 2.016, then the molar mass of the remaining element is 34.1 2.016 = 32.1 g/mol, which is roughly Sulfur (32.065). So the acid is H 2 S. Chapter 9 Worksheet page 5 of 5