Chapter 7 Reactions in solution

Describing Reactions in Solution Balanced Chemical Equation Reactants and products as compounds AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq) Total Ionic Equation All strong electrolytes shown as ions Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)  AgCl(s) + Na+(aq) + NO3(aq) 4.6

Describing Reactions in Solution Net Ionic Equation Show only components that actually react Ag+(aq) + Cl(aq)  AgCl(s) Na+ and NO3 are spectator ions 4.6

Example Balanced Chemical Equation Pb(NO3)2(aq) + 2KI(aq)  2KNO3(aq) + PbI2(s)

Total ionic equations For all dissolved substances show as constituent ions (except weak acids or other weak electrolytes) All others show as molecular formula Pb(NO3)2(aq) + 2KI(aq)  2KNO3(aq) + PbI2(s) Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) 2K+(aq) + 2NO3-(aq) + PbI2(s)

Net ionic equations Spectator ions are those ions that do not undergo a change; they do not participate in the chemical change and are the same on both sides of the equation Remove all spectator ions from the equation Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) = 2K+(aq) + 2NO3-(aq) + PbI2(s)

Pb2+(aq) + 2I-(aq)  PbI2(s) Net ionic equations Pb2+(aq) + 2I-(aq)  PbI2(s) Mass and charge must still balance, although overall charge may not be neutral in a net ionic equation

Pb2+(aq) + 2I-(aq) = PbI2(s) 7.2 Qualitative Analysis Pb2+(aq) + 2I-(aq) = PbI2(s)

For Example: Ag+ and Sr+2 We try to find some anion which could form a precipitate with only one of our two cations at a time. Assume one or both of these cation is in solution If a precipitate is formed, we can then assume that the ion we are looking for is in fact present; if no precipitate forms, the ion is absent. If a precipitate forms, then we filter it off and add another anion to precipitate the second ion. If a precipitate forms, then the second ion is present. If a precipitate does not form, then the second ion is not present in the solution.

Developing a Qualitative Analysis Scheme For many ionic substances the decision can be made using the solubility table. Looking at the previous example, let’s develop a qualitative analysis scheme.

2. Set up a table of solubilities. Look at the Solubility Table to find anions which can precipitate Ag+ and Sr2+. We can use: Cl-, SO42-, S2-, OH-, and PO43-. 2. Set up a table of solubilities. Cl- SO42- S2- OH- PO43- Ag+ Ppt ppt Sr2+ --- ---- --

Add SO42- or PO43- to try to precipitate Sr2+. Procedure: Start by adding Cl-, S2-, or OH- to try to precipitate Ag+ (we do not use SO42- or PO43- as they could also ppt Sr2+). If a precipitate forms, then there is Ag+ present. Filter off and discard the precipitate. Keep the left over solution for the next part. Add SO42- or PO43- to try to precipitate Sr2+. If a precipitate forms, then there is Sr2+ present.

Practice Exercises Draw a qualitative analysis scheme showing how you would separate a mixture of Mg+2, Pb+2, and Sr+2 ions. Cl- SO42- S2- OH- PO43- Mg+2 Pb+2 Sr+2

Practice Exercises Draw a qualitative analysis scheme showing how you would separate a mixture of Mg+2, Pb+2, and Sr+2 ions. Cl- SO42- S2- OH- PO43- Mg+2 ppt Pb+2 Sr+2

Stoichiometry and Reactions in Solution Mass given required Volume of Given Solution Volume of Required Solution mol required mol given

Stoichiometry and Reactions in Solution Mass given required Volume of Given Solution Volume of Required Solution mol required mol given Pb(NO3)2(aq) + 2KI (aq)  PbI2(s) + 2 KNO3(aq) What volume of 4.0 M KI solution is required to yield 89 g PbI2?

Stoichiometry for Reactions in Solution Step 1) Identify the species present in the combined solution, and determine what reaction occurs. Step 2) Write the balanced net ionic equation for the reaction. Step 3) Calculate the moles of reactants. Step 4) Determine which reactant is limiting. Step 5) Calculate the moles of product or products, as required. Step 6) Convert to grams or other units, as required. Stoichiometry steps for reactions in solution

Solution stoichiometry How much volume of one solution to react with another solution Given volume of A with molarity MA Determine moles A Determine moles B Find target volume of B with molarity MB

1 Pb(NO3)2(aq) + 2 KI (aq)  1 PbI2(s) + 2 KNO3(aq) ? L 4.0 M 89 g What volume of 4.0 M KI solution is required to yield 89 g PbI2? Strategy: (1) Find mol KI needed to yield 89 g PbI2. (2) Based on (1), find volume of 4.0 M KI solution. 1 mol PbI2 2 mol KI X mol KI = 89 g PbI2 = 0.39 mol KI 461 g PbI2 1 mol PbI2 M = mol L L = mol M = 0.39 mol KI 4.0 M KI = 0.098 L of 4.0 M KI

How many mL of a 0.500 M CuSO4 solution will react w/excess Al to produce 11.0 g Cu? Al3+ SO42– __CuSO4(aq) + __Al (s)  __Cu(s) + __Al2(SO4)3(aq) 3 CuSO4(aq) + Al (s)  Cu(s) + Al2(SO4)3(aq) 2 3 1 x mol 11 g 1 mol Cu 3 mol CuSO4 X mol CuSO4 = 11 g Cu = 0.173 mol CuSO4 63.5 g Cu 3 mol Cu M = mol L L = mol M 0.173 mol CuSO4 0.500 M CuSO4 = 0.346 L 1000 mL 0.346 L = 346 mL 1 L

Stoichiometry Problems How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3? Cu + 2AgNO3  2Ag + Cu(NO3)2 ? g 1.5L 0.10M 1.5 L .10 mol AgNO3 1 L 1 mol Cu 2 mol AgNO3 63.55 g Cu 1 mol Cu = 4.8 g Cu Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

The Titration of an Acid with a Base

Titration Use a solution of known concentration of base to determine concentration of an unknown of acid (or vice versa) Must be able to identify endpoint of titration to know stoichiometry: Use a pH indicator

Key Titration Terms Titrant – solution of known concentration used in titration Analyte – substance being analyzed Equivalence point – enough titrant added to react exactly with the analyte Endpoint – the indicator changes color so you can tell the equivalence point has been reached 4.8

Measuring Pipets and Volumetric Pipets Measure Liquid Volume

ACID-BASE REACTIONS Titrations H2C2O4(aq) + 2 NaOH(aq) ---> acid base Na2C2O4(aq) + 2 H2O(liq) Carry out this reaction using a TITRATION.

Setup for titrating an acid with a base

Sample Problem In this sample titration, we are trying to determine the concentration of 20.00 mL of HCl. In the titration we will be neutralizing the HCl with 0.150 M NaOH. The volume of base used is 26.23, 26.14, and 26.32 mL for 3 trials. What is the concentration of the acid?

Step 1: The NaOH, the titrant, is placed in the buret Step 1: The NaOH, the titrant, is placed in the buret. The titrant is the solution of known concentration that is added from the buret. Step 2: The HCl is placed in the Erlenmeyer flask and 2-3 drops of phenolphthalein indicator. Since the solution in the flask is acidic, phenolphthalein is colourless. Step 3: NaOH is slowly added to the HCl in the flask.

The equivalence point of the titration is reached when equal numbers of moles of hydrogen and hydroxide ions have been reacted. When this happens in this titration, the pH of the solution in the flask is 7.0 and the phenolphthalein is about to change colour.

Step 4: Add as little excess NaOH as possible Step 4: Add as little excess NaOH as possible. We want to add a single drop of NaOH to the colourless solution in the flask and have the solution in the flask turn pink and stay pink while the contents of the flask are swirled. This permanent colour change in the indicator is known as the endpoint of the titration and the titration is over. Titration Curve

Solve the problem 1st write the equation for the reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) 2nd solve for the amount of moles of the titrant used. NaOH mol = 0.150 mol/L x 0.02567 L= 3.85 x 10-3 mol NaOH Found in titration experiment 3rd using stoichiometry, solve for the concentration of HCl , knowing it is a 1:1 mole ratio 3.85 x 10-3 mol= 0.192 M 0.02000 L

Note Moles Acid= Moles Base n acid=nbase CbaseV acid= CbaseVbase

SAMPLE PROBLEM 2 In an acid-base titration, 17.45 mL of 0.180 M nitric acid, HNO3, were completely neutralized by 14.76 mL of aluminium hydroxide, Al(OH)3. Calculate the concentration of the aluminium hydroxide.

SAMPLE ANSWER 2 The balanced equation for the reaction is: 3HNO3(aq) + Al(OH)3(aq) → Al(NO3)3(aq) + 3H2O(l) The number of moles of nitric acid used is: y mol = 0.180 mol/L x 0.01745 L = 3.14 x 10-3 mol HNO3 From the stoichiometry of the reaction, the number of moles of aluminium hydroxide reacted is: 3.14 x 10-3 mol HNO3 x 1 mol Al(OH)3 = 1.05 x 10-3 mol 3 mol HNO3 Therefore, the concentration of the aluminium hydroxide is: 1.05 x 10-3 mol Al(OH)3 = 0.0711 M 0.01476 L