CH 115 Fall 2014 Worksheet 26 1. What is a solution composed of? Solute – substance you are dissolving Solvent – what you are dissolving the solute in 2. What is the formula for calculating molarity of a solution? Molarity = moles of solute / liters of solution 3. Complete the following calculations. a). What is the molarity of a solution made by dissolving 2.5 g of NaCl in enough water to make 125 ml of solution? 2.5 g NaCl * (1 mol NaCl / 58.45 g NaCl) = 0.0427715997 moles NaCl 125 mL (1 liter / 1000 mL) = .125 L Molarity = moles/L = 0.0427715997 moles/.125 L = 0.342 M b). How many moles of salt are contained in 300 mL of a 0.40 M NaCl solution? M = mol/L mol = M * V Moles = 0.40 M * .300 L = 0.12 moles salt c). Battery acid is generally 3 M H2SO4. Roughly how many grams of H2SO4 are in 400 mL of this solution? M = mol/L mol = M*V Mol = 3 M * .400 L = 1.2 moles H2SO4 * (98 g H2SO4 / 1 mol H2SO4) = 117.6 g d). A chemist dissolves 98.4 g of FeSO4 in enough water to make 2.000 L of solution. What is the molarity of the solution? 98.4 g FeSO4 * (1 mol FeSO4 / 151.85 g FeSO4) = 0.6480079025 mol FeSO4 M = 0.6480079025 mol / 2 L = 0.324 M 4. You have a stock solution that is 0.276 M. How much volume of this stock solution do you need to add to a solution to make 100 mL of 0.128 M solution? M1V1 = M2V2 V1 = M2V2 / M1 = (0.128 M)(100 mL) / 0.276 M = 46.38 mL 5. Write the balanced equation for the reaction of calcium carbonate with hydrochloric acid. What mass of calcium carbonate is required to react with 25 mL of 0.75 M of hydrochloric acid? CaCO3 + 2 HCl CO2 + H2O + CaCl2 .025 L * 0.75 M = 0.01875 mol HCl * (1 mol CaCO3 / 2 mol HCl) * (100.08 g CaCO3 / 1 mol CaCO3) = 0.938 g CaCO3 6. What is a titration? What is the difference between the end point and the equivalence point of a titration? What is special about the equivalence point? A titration is a controlled experiment involving an acid-base neutralization reaction. The end point of a titration is the point at which the indicator CH 115 Fall 2014 Worksheet 26 changes colors. The equivalence point of a titration is the point at which the moles of acid in the experiment equal the moles of base (M1V1 = M2V2). 7. A titration reveals that 11.6 mL of 3.0 M sulfuric acid are required to neutralize the sodium hydroxide in 25.00 mL of NaOH solution. What is the molarity of the NaOH solution? H2SO4 + 2 NaOH 2 H2O + Na2SO4 0.0116 L * 3.0 M = 0.0348 moles H2SO4 * (2 moles NaOH / 1 mole H2SO4) = 0.0696 moles NaOH M = moles/L = 0.0696 moles/ .025 L = 2.78 M