# 1 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Reactions in Solution Sections 5.8-5.10.

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1 © 2006 Brooks/Cole - Thomson Quantitative Aspects of Reactions in Solution Sections 5.8-5.10

2 © 2006 Brooks/Cole - Thomson TerminologyTerminology In solution we need to define the SOLVENTSOLVENT the component whose physical state is preserved when solution forms; usually the component in the largest proportion SOLUTESOLUTE the other solution component

3 © 2006 Brooks/Cole - Thomson Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Concentration (M) = [ …]

4 © 2006 Brooks/Cole - Thomson 1.0 L of water was used to make 1.0 L of solution. Notice the water left over. CCR, page 206

5 © 2006 Brooks/Cole - Thomson Preparing a Solution Active Figure 5.18

6 © 2006 Brooks/Cole - Thomson Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent.Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.Dilute a concentrated solution to give one that is less concentrated.

7 © 2006 Brooks/Cole - Thomson PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

8 © 2006 Brooks/Cole - Thomson The Nature of a CuCl 2 Solution: Ion Concentrations CuCl 2 (aq) --> Cu 2+ (aq) + 2 Cl - (aq) If [CuCl 2 ] = 0.30 M, then [Cu 2+ ] = 0.30 M [Cl - ] = 2 x 0.30 M

9 © 2006 Brooks/Cole - Thomson Step 1: Calculate moles of acid required. (0.0500 mol/L)(0.250 L) = 0.0125 mol Step 2: Calculate mass of acid required. (0.0125 mol )(90.00 g/mol) = 1.13 g USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

10 © 2006 Brooks/Cole - Thomson Preparing a Solution by Dilution

11 © 2006 Brooks/Cole - Thomson PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!

12 © 2006 Brooks/Cole - Thomson PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution

13 © 2006 Brooks/Cole - Thomson PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M V = M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH)(1 L/0.50 mol) = 0.30 L or 300 mL

14 © 2006 Brooks/Cole - Thomson PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.

15 © 2006 Brooks/Cole - Thomson A shortcut A shortcut C initial V initial = C final V final Preparing Solutions by Dilution

16 © 2006 Brooks/Cole - Thomson pH, a Concentration Scale pH: a way to express acidity -- the concentration of H + in solution. Low pH: high [H + ] High pH: low [H + ] Acidic solutionpH < 7 Neutral pH = 7 Neutral pH = 7 Basic solution pH > 7 Basic solution pH > 7 Acidic solutionpH < 7 Neutral pH = 7 Neutral pH = 7 Basic solution pH > 7 Basic solution pH > 7

17 © 2006 Brooks/Cole - Thomson The pH Scale pH = log (1/ [H + ]) = - log [H + ] In a neutral solution, [H + ] = [OH - ] = 1.00 x 10 -7 M at 25 o C pH = - log [H + ] = -log (1.00 x 10 -7 ) = - [0 + (-7)] = 7

18 © 2006 Brooks/Cole - Thomson [H + ] and pH If the [H + ] of soda is 1.6 x 10 -3 M, the pH is ____? Because pH = - log [H + ] then pH= - log (1.6 x 10 -3 ) pH= - log (1.6 x 10 -3 ) pH = -{log (1.6) + log (10 -3 )} pH = -{0.20 - 3.00) pH = 2.80

19 © 2006 Brooks/Cole - Thomson pH and [H + ] If the pH of Coke is 3.12, it is ____________. Because pH = - log [H + ] then log [H + ] = - pH log [H + ] = - pH Take antilog and get [H + ] = 10 -pH [H + ] = 10 -3.12 = 7.6 x 10 - 4 M

20 © 2006 Brooks/Cole - Thomson Zinc reacts with acids to produce H 2 gas.Zinc reacts with acids to produce H 2 gas. Have 10.0 g of ZnHave 10.0 g of Zn What volume of 2.50 M HCl is needed to convert the Zn completely? What volume of 2.50 M HCl is needed to convert the Zn completely? SOLUTION STOICHIOMETRY Section 5.10

21 © 2006 Brooks/Cole - Thomson GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass zinc Stoichiometric factor Moles zinc Moles HCl Mass HCl Volume HCl

22 © 2006 Brooks/Cole - Thomson Step 1: Write the balanced equation Zn(s) + 2 HCl(aq) --> ZnCl 2 (aq) + H 2 (g) Step 2: Calculate amount of Zn Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 3: Use the stoichiometric factor

23 © 2006 Brooks/Cole - Thomson Step 3: Use the stoichiometric factor Zinc reacts with acids to produce H 2 gas. If you have 10.0 g of Zn, what volume of 2.50 M HCl is needed to convert the Zn completely? Step 4: Calculate volume of HCl req’d

24 © 2006 Brooks/Cole - Thomson ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

25 © 2006 Brooks/Cole - Thomson Setup for titrating an acid with a base Active Figure 5.23

26 © 2006 Brooks/Cole - Thomson Titration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3. Indicator shows when exact stoichiometric reaction has occurred. 4. Net ionic equation H + + OH - --> H 2 O H + + OH - --> H 2 O 5. At equivalence point moles H + = moles OH - moles H + = moles OH -

27 © 2006 Brooks/Cole - Thomson 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentra-tion of the NaOH? 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentra-tion of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.

28 © 2006 Brooks/Cole - Thomson 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H 2 C 2 O 4 Step 2: Calculate amount of NaOH req’d

29 © 2006 Brooks/Cole - Thomson 1.065 g of H 2 C 2 O 4 (oxalic acid) requires 35.62 mL of NaOH for titration to an equivalence point. What is the concentration of the NaOH? Step 1: Calculate amount of H 2 C 2 O 4 = 0.0118 mol acid = 0.0118 mol acid Step 2: Calculate amount of NaOH req’d = 0.0236 mol NaOH = 0.0236 mol NaOH Step 3: Calculate concentration of NaOH [NaOH] = 0.663 M

30 © 2006 Brooks/Cole - Thomson LAB PROBLEM #2: Use standardized NaOH to determine the amount of an acid in an unknown. Apples contain malic acid, C 4 H 6 O 5. C 4 H 6 O 5 (aq) + 2 NaOH(aq) ---> Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) Na 2 C 4 H 4 O 5 (aq) + 2 H 2 O(liq) 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid?

31 © 2006 Brooks/Cole - Thomson 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: Calculate amount of NaOH used. C V = (0.663 M)(0.03456 L) = 0.0229 mol NaOH = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated. = 0.0115 mol acid

32 © 2006 Brooks/Cole - Thomson 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 3: Calculate mass of acid titrated. Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid = 0.0115 mol acid

33 © 2006 Brooks/Cole - Thomson 76.80 g of apple requires 34.56 mL of 0.663 M NaOH for titration. What is weight % of malic acid? Step 1: Calculate amount of NaOH used. = 0.0229 mol NaOH = 0.0229 mol NaOH Step 2: Calculate amount of acid titrated = 0.0115 mol acid = 0.0115 mol acid Step 3: Calculate mass of acid titrated. = 1.54 g acid = 1.54 g acid Step 4: Calculate % malic acid.