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Solutions and Stoichiometry

Solutions and Stoichiometry. Many times the reactants and/or products of chemical reactions are water solutions. In these cases, the concentration of the solution must be determined in order to determine amounts of reactants or products

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Solutions and Stoichiometry

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  1. Solutions and Stoichiometry • Many times the reactants and/or products of chemical reactions are water solutions. • In these cases, the concentration of the solution must be determined in order to determine amounts of reactants or products • The concentration of a solution is a measure of the amount of solute that is dissolved in a given amount of solvent 1

  2. Solutions and Stoichiometry • A solution is made up of a Solute and a Solvent • A Solute is what is being dissolved. • A Solvent is what does the dissolving • (the most “stuff” present) • Note: not all solutions are Liquids • Air (O2 in N2)(usually measure in PPM or PPB). • Alloys • (metals mixed with other metals) Steel • Brass (copper and zinc) • Bronze (copper and tin) 2

  3. Concentration • For individual elements, compounds and molecules (solid & liquids): Density (g/mL) • For solutions: • Molarity (M) • (moles/liter) (“Molar”) • Molality (m) • (moles/kilograms) (“Molal”) • (rarely used) 3

  4. Molarity • The most common concentration unit is Molarity 4

  5. Molarity Calculations How many grams of NaOH are required to prepare 250 cm3 of 0.500 M solution? • Molar Mass of NaOH = 22.99+16+1.01 = 40.00 g/mol • 250 cm3 (mL) = 0.250 dm3 (L) 5

  6. Molarity Calculations Calculate the concentration of a NaCl solution that contains 24.5 g of NaCl in 250 cm3 of solution. • Molar mass of NaCl = 23.0 + 35.5 = 58.5 6

  7. Copper metal reacts with nitric acid according to the following reaction: 8 HNO3 (aq) + 3 Cu  3 Cu(NO3)2 (aq) + 4 H2O (l) + 2 NO (g) What volume of 8.00 M HNO3 would be required to consume a copper penny whose mass is 3.08 grams? Stoichiometry Calculations Involving Solutions 1 = 16.2 cm3 HNO3 7

  8. 15.0 cm3 of a 0.500 M AgNO3 solution is required to precipitate the sodium chloride in 10 cm3 of a salt solution. What is the concentration of the solution? AgNO3 (aq)+ NaCl(aq)AgCl (s) +NaNO3 (aq) Molar Mass NaCl = 23.0 + 35.5 = 58.5 g/mol Stoichiometry Calculations Involving Solutions 2 = 0.0075 mol NaCl 0.0075 mol NaCl 1 0.0100 dm3 = 0.75 mol dm-3 or 0.75 M 8

  9. If 15.0 cm3 of a 0.500 M AgNO3 solution is added to 33.0 cm3 of a 1.875 M AgNO3solution.What is the concentration of the solution? Solutions Concentrations = 0.0075 molAgNO3soln 1 = 0.0619 molAgNO3soln 2 New conc = new moles/new volume: 0.0075 mol + 0.0619 mol =0.0694 moles AgNO3 (new moles) 0.0150 dm3 + 0.0330 dm3 = 0.0480 dm3 (new volume) 0.0694 moles/0.0480 dm3 = 1.45mol dm-3or1.45 M AgNO3 9

  10. Sometimes, solutions of lesser concentration need to be made. This is done by adding solvent. Using the equation, M1 V1 = M2 V2we can calculate a volume or molarity of solutions. M1 is the Molarity of the starting solution V1 is the Volume of the starting solution M2 is the Molarity of the end solution V2 is the Volume of the end solution Knowing any 3 variables, we can calculate the last variable (algebra). Dilutions 10

  11. Using the equation, M1 V1 = M2 V2 What is the final molarity of NaOH solution if 10 cm3 of 3.5 M NaOH solution is diluted to 250 cm3? M1 = 3.5 M V1 = 10 cm3 M2= ? V2 = 250 cm3 M1V1 = M2 V2 3.5M x 10 cm3 = M2x 250 cm3 M2= 3.5M x 10 cm3 250 cm3 M2 = 0.14 M NaOH Dilutions 11

  12. Using the equation, M1 V1 = M2 V2 How much water must be added to change 150 cm3 of 2.5 M HNO3 to a 0.75M solution? M1 = 2.5 M V1 = 150 cm3 M2 = 0.75 M V2 = ? cm3 M1V1 = M2 V2 2.5 M x 150 cm3 = 0.75 M x V2 V2 = 2.5M x 150 cm3 0.75 M V2 = 500 cm3 BUT remember, that’s the TOTAL volume. 500 cm3 – 150 cm3 = 350 cm3 water added. Dilutions 12

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