Basic Chemistry Copyright © 2011 Pearson Education, Inc Molarity and Dilution Chapter 12 Solutions

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 2 Molarity (M) Molarity (M) is a concentration term for solutions the moles of solute in 1 L of solution moles of solute liter of solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 3 Preparing a 6.0 M Solution A 6.00 M NaOH solution is prepared by weighing out 60.0 g of NaOH (1.50 mol) and adding water to make L of a 6.00 MNaOH solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 4 Calculating Molarity

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 5 What is the molarity of L of a NaOH solution if it contains 6.00 g of NaOH? STEP 1 State the given and needed quantities. Given 6.00 g of NaOH in L of solution Need molarity (M) STEP 2 Write a plan to calculate molarity. molarity (M) = moles of solute liters of solution grams of NaOH moles of NaOH molarity Example of Calculating Molarity

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 6 STEP 3 Write equalities and conversion factors needed. 1 mol of NaOH = g of NaOH 1 mol NaOH and g NaOH g NaOH 1 mol NaOH STEP 4 Set up problem to calculate molarity g NaOH x 1 mol NaOH = mol of NaOH g NaOH mol NaOH = mol L solution 1 L = M NaOH solution Example of Calculating of Molarity (continued)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 7 What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? A M NaHCO 3 solution B M NaHCO 3 solution C M NaHCO 3 solution Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 8 STEP 1 State the given and needed quantities. Given 46.8 g of NaHCO 3 in L of solution Need molarity (M) STEP 2 Write a plan to calculate molarity. molarity (M) = moles of solute liters of solution grams of NaHCO 3 moles of NaHCO 3 molarity Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 9 STEP 3 Write equalities and conversion factors needed. 1 mol of NaHCO 3 = g of NaHCO 3 1 mol NaHCO 3 and g NaHCO g NaOH 1 mol NaHCO 3 STEP 4 Set up problem to calculate molarity g NaHCO 3 x 1 mol NaHCO 3 = mol of NaHCO g NaHCO mol NaHCO 3 = 1.71 M NaHCO 3 solution L Solution (continued)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 10 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions.

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 11 Example of Using Molarity in Calculations How many grams of KCl are needed to prepare L of a M KCl solution? STEP 1 State the given and needed quantities. Given L of a M KCl solution Need grams of KCl STEP 2 Write a plan to calculate mass or volume. liters of KCl solution moles of KCl grams of KCl

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 12 Example of Using Molarity in Calculations (continued) STEP 3 Write equalities and conversion factors needed. 1 mol of KCl = g of KCl 1 mol KCl and g KCl g KCl 1 mol KCl 1 L of KCl solution = mol of KCl 1 L KCl solution and mol KCl mol KCl 1 L KCl solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 13 Example of Using Molarity in Calculations (continued) STEP 4 Set up problem to calculate mass or volume L x mol KCl x g KCl = 6.71 g of KCl 1 L 1 mol KCl

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 14 How many grams of AlCl 3 are needed to prepare 37.8 mL of a M AlCl 3 solution? A g of AlCl 3 B g of AlCl 3 C g of AlCl 3 Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 15 Solution STEP 1 State the given and needed quantities. Given 37.8 mL of a M AlCl 3 solution Need grams of AlCl 3 STEP 2 Write a plan to calculate mass or volume. milliliters of AlCl 3 solution liters of AlCl 3 solution moles of AlCl 3 grams of AlCl 3

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 16 Solution (continued) STEP 3 Write equalities and conversion factors needed. 1 mol of AlCl 3 = g of AlCl 3 1 mol AlCl 3 and g AlCl g AlCl 3 1 mol AlCl mL of AlCl 3 solution = 1 L of AlCl 3 solution 1000 mL AlCl 3 solution and 1 L AlCl 3 solution 1 L AlCl 3 solution 1000 mL AlCl 3 solution 1 L of AlCl 3 solution = mol of AlCl 3 1 L AlCl 3 solution and mol AlCl mol AlCl 3 1 L AlCl 3 solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 17 Solution (continued) STEP 4 Set up problem to calculate mass or volume mL x 1 L x mol x g 1000 mL 1 L 1 mol = g of AlCl 3 (B)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 18 How many milliliters of a 2.00 M HNO 3 solution contain 24.0 g of HNO 3 ? A mL of HNO 3 solution B mL of HNO 3 solution C mL of HNO 3 solution Learning Check

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 19 STEP 1 State the given and needed quantities. Given 24.0 g of HNO 3 ; 2.00 M HNO 3 solution Need milliliters of HNO 3 solution STEP 2 Write a plan to calculate mass or volume. g of solution moles of HNO 3 mL of HNO 3 Solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 20 STEP 3 Write equalities and conversion factors needed. 1 mol of HNO 3 = g of HNO 3 1 mol HNO 3 and g HNO g HNO 3 1 mol HNO mL of HNO 3 = 2.00 mol of HNO mL HNO 3 and 2.00 mol HNO mol HNO mL HNO 3 Solution (continued)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 21 STEP 4 Set up problem to calculate mass or volume g HNO 3 x 1 mol HNO 3 x 1000 mL g HNO mol HNO 3 = 190. mL of HNO 3 solution (C) Solution (continued)

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 22 Dilution In a dilution, water is added volume increases concentration decreases

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 23 Comparing Initial and Diluted Solutions In the initial and diluted solution, the moles of solute are the same the concentrations and volumes are related by the equation M 1 V 1 = M 2 V 2 initial diluted

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 24 Calculating Dilution Quantities

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 25 Example of Dilution Calculations What is the final molarity of the solution when L of M KOH is diluted to L? STEP 1 Prepare a table of the initial and diluted volumes and concentrations. Initial Solution Diluted Solution M 1 = M M 2 = ? V 1 = L V 2 = L

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 26 Example of Dilution Calculations (continued) STEP 2 Solve the dilution expression for the unknown quantity. M 1 V 1 = M 2 V 2 V 2 V 2 M 2 = M 1 V 1 V 2 STEP 3 Set up the problem by placing known quantities in the dilution expression. M 2 = M 1 V 1 = (0.600 M)(0.180 L) = M V L

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 27 Learning Check What is the final volume, in milliliters, if 15.0 mL of a 1.80 M KOH solution is diluted to give a M KOH solution? A mL of M KOH solution B mL of M KOH solution C mL of M KOH solution

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 28 Solution STEP 1 Prepare a table of the initial and diluted volumes and concentrations. Initial Solution Diluted Solution M 1 = 1.80 M V 1 = 15.0 mL M 2 = 0.300M V 2 = ?

Basic Chemistry Copyright © 2011 Pearson Education, Inc. 29 Solution (continued) STEP 2 Solve the dilution expression for the unknown quantity. M 1 V 1 = M 2 V 2 M 2 M 2 V 2 = M 1 V 1 M 2 STEP 3 Set up the problem by placing known quantities in the dilution expression. V 2 = M 1 V 1 = (1.80 M)(15.0 mL) M M = 90.0 mL (C )