CHE Chapter Chapter Aqueous Reactions and Solution Stoichiometry. The total concentration of ions in a 0.50 M solution of HCl is. (a)..000 M (b). 0.5 M (c). 0.500 M (d). 0.065 M Explanation: Since HCl is a strong acid, it will dissociate completely in solution according to the equation: HCl H Cl -. Since the mole:mole ratio between the HCl and the H and Cl - ions is :, a 0.50 M solution on dissociation will produce 0.50 M solution of H and 0.50 M solution of Cl -. Thus the sum of thee two will be equal to 0.500 M. A strong electrolyte is one that completely in solution. (a). reacts (b). decomposes (c). disappears (d). ionizes 3. A weak electrolyte exists predominantly as in solution. (a). atoms (b). ions (c). free radicals (d). molecules. Which of the following are strong electrolytes? HC, HC H 3 O, NH 3, KC (a). HC, NH 3, NaC (b). HC, NaC (c). HC, HC H 3 O, NH 3, NaC (d). HC, HC H 3 O, NaC Explanation: HCl is a strong electrolyte by definition while since NaCl is a water soluble ionic salt it will dissolve completely, making it a strong electrolyte also. Copyright 006 Dr. Harshavardhan D. Bapat
CHE Chapter 5. Which of the following are weak electrolytes? HC, HC H 3 O, NH 3 (a). HC, NH 3 (b). HC H 3 O, HC (c). HC H 3 O, NH 3 (d). All of these 6. Which of the following are strong acids? HI, HNO 3, HF, HBr (a). HF, HBr (b). HI, HNO 3, HBr (c). HI, HF, HBr (d). HNO 3, HF, HBr 7. Which hydroxides are strong bases? Sr(OH), KOH, NaOH, Ba(OH) (a). KOH, Ba(OH) (b). KOH, NaOH (c). KOH, NaOH, Ba(OH) (d) Sr(OH), KOH, NaOH and Ba(OH) 8. A neutralization reaction between an acid and a metal hydroxide produces. (a). water and a salt (b). hydrogen gas (c). precipitate (d). a sodium salt 9. What is the concentration (M) of KC in a solution made by mixing 5.0 ml of 0.00 M KC with 50.0 ml of 0.00 M KC? (a). 0.0500 (b). 0.00 (c). 0.0333 (d). 0.050 Explanation: Since the solutions have the same composition and concentration to begin with, it will not change the concentration of the final 75.0 ml of solution. Copyright 006 Dr. Harshavardhan D. Bapat
CHE Chapter 0. What is the concentration (M) of CH 3 OH in a solution prepared by dissolving.7 g of CH 3 OH in sufficient water to give exactly 30 ml of solution? (a)..7 (b)..30 x 0 - (c). 0.08 (d)..59 Explanation: Need to convert the grams of CH 3 OH to moles and then find the molarity of the solution by using the molarity formula. Do not forget to convert the ml to L. mole CH 3OH.7 g CH 3 OH =.59 M 3.0 g 0.30 L. How many grams of H 3 PO are in 35. ml of a.75 M solution of H 3 PO? (a). 0.6 (b). 9.6 (c). 0 (d)..9 Explanation: Need to convert the ml of H 3 PO to liters and then find the # of moles of phosphoric acid. The moles of phosphoric acid can then be converted to grams of phosphoric acid. L 97.99 g H3PO 35. ml.75 M = 9.6 g H3PO 000 ml mole. What is the concentration (M) of a Na SO solution prepared by dissolving 5.35 g of Na SO in sufficient water to give 330 ml of solution? (a).. x 0 (b). 0.06 (c). 6.7 (d). 0. Explanation: Convert grams of Na SO to moles of Na SO, ml of water to liters of water and then find the molarity of the solution by using the molarity formula. mole Na SO 5.35 g Na = SO 0. M Na SO.035 g Na SO 0.330 L Copyright 006 Dr. Harshavardhan D. Bapat 3
CHE Chapter 3. How many grams of LiOH are there in 750.0 ml of a 0.058 M LiOH solution? (a).. x 0-5 (b)..3 (c). 0.8 (d). 3.50 Explanation: Calculate the number of moles of LiOH present in this solution using the molarity formula and then convert the number of moles to grams of LiOH. 3.98 g 7.50 0 " L 0.058 M = 0.8 g LiOH mole LiOH. How many moles of the iodide ions are present in 5.6 ml of a.00 x 0-3 M solution of Cobalt (II) iodide? (a)..00 (b). 0.500 (c). 0.60 (d). 3.65 x 0 - Explanation: When CoI dissolves in water it will ionize completely to produce cobalt and iodide ions according to the equation: CoI Co I -, producing moles of iodide ions per CoI mole. Calculate the number of moles of CoI present in the solution first and then factor in the stoichiometric ratio between the iodide and the CoI. 3 moles I.56" 0 L ".00" 0 M " = 3.65" 0 moles iodide mole CoI 5. How many moles of K are present in 3.3 ml of a.7 M solution of K 3 PO? (a). 0.36 (b). 0.3 (c)..7 (d). 0.033 Explanation: When K 3 PO dissolves in water it will ionize completely according to the equation: K 3 PO 3K PO 3-, producing 3 moles of K ions every mole of K 3 PO. Calculate the number of moles of K 3 PO present and then factor in the stoichiometric ratio between potassium and the K 3 PO. 3.3 3 moles K mole K PO " 0 L ".7 M " = 0.3 moles of K 3 Copyright 006 Dr. Harshavardhan D. Bapat
CHE Chapter 6. What are the respective concentrations (M) of Na and SO - produced by dissolving 0.500 mol Na SO in water and diluting to.33 L? (a). 0.665 and 0.665 (b). 0.665 and.33 (c). 0.75 and 0.376 (d) 0.376 and 0.75 Explanation: Calculate the molarity of the solution and then the molarities of individual ions by factoring in the stoichiometric ratios between each ion and the sodium sulfate. When sodium sulfate dissolves in water each mole of sodium sulfate will generate moles of sodium and one mole of sulfate ions. Na SO Na SO - 0.500 moles Na 0.500 moles Na SO SO moles Na mole Na SO - moleso mole Na SO.33 L.33 L = 0.75 moles Na = 0.376 moles SO 7. How many ml of 0.875 M NaOH will be required to completely titrate 5.00 ml of 0.870 M HCl? (a) 5.00 ml (b).9 ml (c) 9.9 ml (d) None of the above Explanation: Do not use the M V = M V formula here. Need to write a balanced equation for the reaction between NaOH and HCl and then calculate the ml of NaOH needed by using the stoichiometric relationships between the reactants and the molarity formula. - HCl NaOH 5.00 ml?? 0.870 M 0.875 M NaCl H O molenaoh.5 0 " LHCl 0.870MHCl molehcl =.9mlNaOH L 0.875MNaOH 000ml L 8. How many ml of 0.875 M NaOH will be required to completely titrate 5.00 ml of 0.870 M H SO? (a) 5.00 ml (b) 9.9 ml (c) 9.7 ml (d) None of the above Copyright 006 Dr. Harshavardhan D. Bapat 5
CHE Chapter Explanation: Do not use the M V = M V formula here. Need to write a balanced equation for the reaction between NaOH and H SO and then calculate the ml of NaOH needed by using the stoichiometric relationships between the reactants and the molarity formula. NaOH H SO Na SO H O.50 0 " molesnaoh 000ml L 0.870MH SO = 9. 7ml moleh SO 0.875M L 9. The molarity of a solution prepared by diluting 3.7 ml of 5.005 M aqueous K Cr O 7 to 500 ml is. (a). 57. (b). 0.00 (c). 0.38 (d). 0.0879 Explanation: You should use the M V = M V formula here. It is the only formula that you should use here since this is just a dilution of the same substance and no chemical reactions are involved. Change the volumes to liters on both sides to get the correct answer. 0. Consider the following 0.00 M solutions. Which one of these solutions will have the highest concentration of sodium ions? (a). Sodium phosphate (b). Sodium sulfate (c). Sodium hydroxide (d). Sodium nitrate Explanation: Knowing the formulas of these compounds allows you to find the answer here. The formulas are Na 3 PO, Na SO, NaOH and NaNO 3 respectively and the Na 3 PO has the largest moles of sodium ions per mole of compound compared to all others. Copyright 006 Dr. Harshavardhan D. Bapat 6