Solution Chemistry Practice Problems – The Answers

Solution Chemistry Practice Problems – The Answers 1. How would you make 500.0 mL of 3.0 M HNO3 solution if all you had on hand were distilled water and 16.0 M HNO3? 16.0 M is the original solution (C1) with 500.0 mL (V2) and 3.0 M (C2). (16.0 M x V1) = (3.0 M x 500.0 mL) V1 = 93.75 mL or 94 mL of 16.0 M nitric acid is added to water and then diluted to 500.0 mL with distilled water and mixed. 2. How would you make 600.0 mL of a 2.00 M calcium hydroxide solution from solid calcium hydroxide? Use formula M x V = moles where moles is sample (mass/molecular mass) 2.00 M x 0.6000 L = moles of calcium hydroxide moles of calcium hydroxide = 1.2 moles calcium hydroxide Calcium hydroxide is Ca(OH)2 its molecular mass is 74.0 g/mol 1.2 mole calcium hydroxide = (x/74 g/mol) x = 88.8 grams of calcium hydroxide is dissolved in distilled water, then diluted with until volume of 600.0 mL and mixed 3. Prepare 750.0 mL of 0.500 M solution of sodium chloride from solid sodium chloride. Use formula M x V = moles where moles is sample (mass/molecular mass) 0.500 M x 0.7500 L = (x/58 g/mol) x = 21.8 grams of NaCl is dissolved in distilled water, then diluted to 750.0 mL and mixed. 4. Research grade hydrogen peroxide is a 30.0 % (w/v) solution. The weight to volume percent (w/v)% is defined as grams of solute per 100.00 mL of solution. Determine the molarity of the 30.0% (w/v) hydrogen peroxide solution. H2O2 is the solute so 30.0% (w/v) means there is 30.0 grams of H2O2 in every 100.0 mL of solution. (30.0 g H2O2/34.0 g/mol) = 0.882 mole H2O2 M is defined as moles of solute/liters of solution 0.882 mole H2O2 / 0.1000 L = 8.82 M H2O2 5. You diluted 50.0 mL of the 30% (w/v) solution of hydrogen peroxide to 125.0 mL, determine the molarity of this diluted solution. 8.82 M H2O2 is (C1) and 50.0 mL (V1) with 125.0 mL (V2) (8.82 M x 50.0 mL) = (125.0 mL x C2) C2 = 3.53 M H2O2 6. A student needs to prepare 250.0 mL of a 1:3 dilution of 0.50 M stock solution of sodium chloride. How would you make 250.0 mL of this solution? A 1: 3 dilution implies that the new molarity of the solution will be 1/3 of the original solution. 0.50 M/3 = 0.17 M. 0.17 M is the concentration of the diluted solution. 0.50 M x V1 = 0.17 M x 250.0 mL So, V1 = 83 mL So to make 250.0 mL of a 1:3 dilution of 0.50 M NaCl, 83 mL will of 0.50 M NaCl will be diluted to 250.0 mL 7. Determine the amount in milliliters of 0.75 M sodium hydroxide titrated to completely precipitate out 1.052 liters of 0.055 M copper (II) chloride solution. 2 NaOH (aq) + CuCl2 (aq)  2 NaCl (aq) + Cu(OH)2 (s) 1.052 L x 0.055M CuCl2 = 0.058 moles CuCl2 (0.058 mol CuCl2) x (1 mol Cu(OH)2/1 mol CuCl2) x (2 mol NaOH/1 mol Cu(OH)2) = 0.12 mol NaOH 0.12 mol NaOH = 0.75 M x V V = 0.16 liter or 160 mL of NaOH is required to precipitate all of the 0.55 M CuCl2 8. What volume of 0.158 M hydrochloric acid is required to neutralize 2.87 grams of magnesium hydroxide? Mg(OH)2 (aq) + 2 HCl  H2O (l) + MgCl2 (aq) (2.87 g Mg(OH)2 / 58.0 g/mol) = 0.0495 mol Mg(OH)2 (0.0495 mol Mg(OH)2) x (2 mol HCl/1 mol Mg(OH)2) = 0.0990 mol HCl 0.0990 mol = 0.158 M x V V = 0.626 L or 626 L of a 0.158 M HCl is needed to neutralize 2.87 grams of Mg(OH) 2 9. A student prepared a potassium hydroxide solution. To determine the actual molarity he standardizes the solution with a primary standard known as potassium hydrogen phthalate (KHP). KHP molecular formula is KHC8H4O4. Determine the molarity of the potassium hydroxide solution the student prepared if it took 25.6 mL of KOH to neutralize 0.32 grams of KHP. KHC8H4O4 (aq) + KOH (aq)  H2O (l) + K2C8H4O4 (aq) (0.32 g KHP/ 204.3 g/mol) = 0.0016 mol KHP 0.0016 mol KHP x (1 mole KO/1 mol KHP) = 0.0016 mol KOH 0.0016 mol KOH/0.0256 L = 0.061M KOH 10. Determine the molarity of a magnesium hydroxide solution given the following information: It took 19.7 mL of 0.085M nitric acid solution to neutralize 15.00 mL of the magnesium hydroxide solution. 2 HNO3 (aq) + Mg(OH)2 (aq)  2 H2O (l) + Mg(NO3)2 (aq) 0.0197 L x 0.085 M = 1.7 x 10-3 mol HNO3 (1.7 x 10-3 mol HNO3) x (1 mol Mg(OH)2 / 2 mol HNO3) = 8.5 x 10-4 mol Mg(OH)2 8.5 x 10-4 mol Mg(OH)2 = 0.01500 L x M M = 5.7 x 10-2 M Mg(OH)2 11. It was determined that only 78.3% of iron (III) hydroxide was precipitated when 175.0 mL of a 0.150 M solution of iron (III) nitrite solution was titrated with 0.200 M calcium hydroxide. a) Determine the mass of iron (III) hydroxide that was actually formed. b) How many milliliters of calcium hydroxide should have been added in order to completely precipitate out all of the iron (III) nitrite solution as iron (III) hydroxide. 2 Fe(NO2)3 (aq) + 3 Ca(OH)2 (aq)  2 Fe(OH)3 (s) + 3 Ca(NO2)2 (aq) a) 0.1755 L x 0.150 M = 0.0264 mol Fe(NO 2)3 0.0264 M Fe(NO2)3 x (2 mol Fe(OH)3 /2 mol Fe(NO2)3) x (106.85 g/1 mol Fe(OH)3) = 2.82 g Percent yield: 0.783 = (x/2.82 g) X = 2.20 grams of Fe(OH)3 was predicated. a) 0.1755 L x 0.150 M = 0.0264 mol Fe(NO 2)3 0.0264 M Fe(NO2)3 x (3 mol Ca(OH)2 /2 mol Fe(NO2)3) = 0.0396 mol Ca(OH)2 0.0396 mol Ca(OH)2 = 0.200 M X V V = 198 mL of 0.200 M Ca(OH)2 should have been added to have a complete precipitation 12. Determine how many grams of sulfuric acid are in 2.00 quarts if 2.50 ml of sulfuric acid requires 34.90 ml of 0.096 M sodium hydroxide to reach the titration’s equivalence point (i.e. neutralization). Useful information: 1 qt. = 0.946 liters H2SO4 (aq) + 2 NaOH (aq)  2 H2O (l) + Na2SO4 (aq) 0.03490 L x 0.96 M = 0.034 mol NaOH 0.034 mol NaOH x (1 mol H2SO4/ 2 mol NaOH) = 0.017 mol H2SO4 0.017 mol H2SO4 = M x 0.00250 L = 6.7 M H2SO4 6.7 M means 6.7 moles of H2SO4 in one liter of solution 6.7 mole H2SO4 x (98.1 g/1 mol H2SO4) = 660 grams H2SO4 in a liter You have 2.00 qt or 2.00 qt x (0.946 L/1 qt) = 1.89 L of H2SO4 (660 grams H2SO4 / 1 L) = (x/1.89 L) 1248.72 grams or 1300 grams of H2SO4 is in 2.00 qt of a 6.7 M H2SO4