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Acids Lesson 16 Titrations
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Calculations to find concentration of sample solution.
Objectives: Calculations to find concentration of sample solution. Calculations to find molar ratio of acid and base in neutralization equation. Practical Aspects of Titration Primary Standards and secondary standards Indicators Titration curves
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Titration
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Titration Purpose: To find concentration of known ion
Method: Acid Base neutralization
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Titration Purpose: To find concentration of know ion
Method: acid base neutralization Material: Burette, Erlenmeyer flask, indicator, measuring cylinder to measure volume, known Concentration base solution, acid solution of unknown concentration
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Titration Procedure: Wash burette and flask with distilled water
Remove water completely from burette and flask Measure 15 mL of acid solution using measuring cylinder and transfer to flask Close tap of burette and fill it to zero
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Titration Procedure: Put 5 drops of indicator solution in flask
Hang the burette into the ring stand. Open the tap of the burette and carefully allow Base solution to drop into the flask drop-wise. Keep stirring the flask to mix acid and base solution.
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Titration Procedure: When colour change is observed then stop and measure volume of base solution that was used to titrate the acid solution
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Titration Calculation
Find moles of the base solution by volume and concentration of standard solution Find moles of sample solution Find concentration of the sample solution by moles and volume of the sample solution
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Titration Terms Standard solution
Solution in burette, known concentration Sample solution Solution in burette, known volume, unknown concentration
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Titration Terms Indicator
A solution that change colour when reaction is complete. This is added into the flask.
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Titration Terms Transition point, end point, stoichiometric point
A point in time during titration process when moles of acid become equal to moles of base
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Setup for titrating an acid with a base
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Titration Calculation
6.50 mL of M H2C2O4 is required to neutralize 10.0 mL of NaOH solution in a titration. Calculate the base concentration. 1H2C2O4 + 2NaOH Na2C2O H2O L L 0.100 M ? M L H2C2O4 x mole 1 L x 2 mole NaOH 1 mole H2C2O4 [NaOH] = L = M
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Calculate the volume of 0. 100 M H2CO3 required to. neutralize 25
Calculate the volume of M H2CO3 required to neutralize 25.0 mL of M KOH. 1H2CO KOH → K2CO HOH ? L L 0.100 M M
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Simple Titration Calculations
Definition of end point, transition point, stoichiometric point Calculations to find % purity of reactants using titration # of protons donated Practical aspects(choice of indicators, unstable standard solution, curves)
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Calculate the volume of 0. 100 M H2CO3 required to. neutralize 25
Calculate the volume of M H2CO3 required to neutralize 25.0 mL of M KOH. 1H2CO KOH → K2CO HOH ? L L 0.100 M M L KOH x mol x 1 mole H2CO3 x L = L L mole KOH mol
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LAB PROBLEM #1: Standardize a solution of NaOH — i. e
LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration. 35.62 mL of NaOH is neutralized with 25.2 mL of M HCl by titration to an equivalence point. What is the concentration of the NaOH?
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2. Calculations to find % purity of reactants using titration
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# of protons donated by acid
3. Calculations to find molar ratio of acid and base in neutralization equation. Or # of protons donated by acid
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2. Calculations to find molar ratio of acid and base in neutralization equation.
One proton donated H3PO KOH → KH2PO4 + H2O Two protons donated H3PO KOH → K2HPO4 + H2O Three protons donated H3PO KOH → K3PO4 + H2O
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# of protons donated by acid
2. Calculations to find molar ratio of acid and base in neutralization equation. Or # of protons donated by acid How can we calculate Which is correct equation(correct molar ratio)? Which salt is being produced?
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0.25M, 20mL H3PO4 reacts with 50mL or 0.20M KOH.
1. How many protons does H3PO4 donates to KOH? 2. What salts is produced? (write formula) One proton donated H3PO KOH → KH2PO4 + H2O Two protons donated H3PO KOH → K2HPO4 + H2O Three protons donated H3PO KOH → K3PO4 + H2O
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0.25M, 20mL H3PO4 reacts with 50mL or 0.20M KOH.
1. How many protons does H3PO4 donates to KOH? 2. What salts is produced? (write formula) H3PO KOH → Moles of H3PO : moles of KOH =0.25M X M : =0.20M X L = : =0.010 = : =2 H3PO KOH → K2HPO4 + H2O Two protons donated
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Two protons donated Salt Produced = K2HPO4
0.25M, 20mL H3PO4 reacts with 50mL or 0.20M KOH. 1. How many protons does H3PO4 donates to KOH? 2. What salts is produced? (write formula) H3PO KOH → Moles of H3PO : moles of KOH =0.25M X M : =0.20M X L = : =0.010 = : =2 H3PO KOH → K2HPO4 + H2O Two protons donated Salt Produced = K2HPO4
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Question 94 On Page 158 Home Work: 95-99
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4. Titration standards
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A standard solution has known molarity.
A primary standard is made by weighing a pure solid and diluting in a volumetric flask. A secondary standard requires a titration to calibrate its concentration. It was made using an acid or base with suspect purity. Not suitable Example Reason Arrhenius bases NaOH Hygroscopic Strong acids HCl volatile liquids or gases Suitable solid bronsted bases Na2CO3 pure solid solid weak acids H2C2O4 pure solid
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Primary Standard Secondary Standard
NaOH(s) Na2CO3(s) H2SO4(l) H2C2O4(s) HCl(g) KC6H5COO(s) C6H5COOH(s) NH3(g)
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Primary Standard Secondary Standard
NaOH(s) Na2CO3(s) H2SO4(l) H2C2O4(s) HCl(g) KC6H5COO(s) C6H5COOH(s) NH3(g)
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Primary Standard Secondary Standard
H2SO4(l) NaOH(s) Na2CO3(s) H2C2O4(s) HCl(g) KC6H5COO(s) C6H5COOH(s) NH3(g)
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Primary Standard Secondary Standard
H2SO4(l) NaOH(s) Na2CO3(s) H2C2O4(s) HCl(g) KC6H5COO(s) C6H5COOH(s) NH3(g)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NaOH(s) Na2CO3(s) HCl(g) KC6H5COO(s) C6H5COOH(s) NH3(g)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NaOH(s) Na2CO3(s) HCl(g) KC6H5COO(s) C6H5COOH(s) NH3(g)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) NaOH(s) Na2CO3(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) NaOH(s) Na2CO3(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) NaOH(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) NaOH(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) NaOH(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) NaOH(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) NaOH(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) NaOH(s) HCl(g) KC6H5COO(s) C6H5COOH(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) C6H5COOH(s) NaOH(s) HCl(g) KC6H5COO(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) C6H5COOH(s) NaOH(s) HCl(g) KC6H5COO(s)
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Primary Standard Secondary Standard
H2SO4(l) H2C2O4(s) NH3(g) Na2CO3(s) C6H5COOH(s) NaOH(s) KC6H5COO(s) HCl(g)
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Preparing a Primary Standard
1. Calculate the mass of H2C2O4.2H2O require to prepare 100.0 mL of M standard solution. L
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Preparing a Primary Standard
1. Calculate the mass of H2C2O4.2H2O require to prepare 100.0 mL of M standard solution. L x mole 1 L
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Preparing a Primary Standard
1. Calculate the mass of H2C2O4.2H2O require to prepare 100.0 mL of M standard solution. L x mole x g 1 L mole
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Preparing a Primary Standard
1. Calculate the mass of H2C2O4.2H2O require to prepare 100.0 mL of M standard solution. L x mole x g = g 1 L mole
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Titration indicators
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The equivalence point is the end of a titration where the stoichiometry of the reaction is exactly satisfied, or moles H+ = moles OH-. The transition point refers to when an indicator changes color and [HInd] = [Ind-].
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Choosing an Indicator When you choose an indicator, you must pick one so that the transition point of the indicator matches the equivalence point of the titration. Rule of thumb Salt Equivalence Point Neutral 7 Basic 9 Acidic 5
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Titration Curves
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Titration Curves A titration curve is a graph of the pH changes that occur during an acid-base titration versus the volume of acid or base added.
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1. Titration Curve: Strong Acid and Strong Base
HCl KOH → KCl HOH neutral salt Indicator pH = 7 Bromothymol Blue- see page 7 50 mL of 0.10 M KOH is added to 25 mL of 0.10 M HCl We start here The pH of 0.10 M HCl is 1.0 The pH of 0.10 M KOH is 13.0 pH Volume .10 M KOH added 14 7 25 50 0.10 M KOH 0.10 M HCl Neutral Salt pH = 7.0
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2. Titration Curve: Weak Acid and Strong Base
HCN KOH → KCN H2O basic salt Indicator pH = 9 Phenolphthalein- see page 7 20 mL of 1.0 M HCN is added to 10 mL of 1.0 M KOH is added to We end here pH = 3 We start here pH = 14 pH Volume 1.0 M HCN added 14 7 1.0 M KOH Basic Salt pH = 9 1.0 M HCN 10 20
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3. Titration Curve: Strong Acid and Weak Base
HCl NH3 → NH4+ + Cl- acid salt Indicator pH = 5 Methyl Red- see page 7 60 mL of 1.0 M NH3 is added to 30 mL of 1.0 M HCl 1.0 M NH3 pH 10 pH Volume 1.0 M NH3 added 14 7 Acid Salt pH 5 30 60 1.0 M HCl pH = 0
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4. Match the Curve with the Reaction
pH Volume .10 M KOH added 14 7 25 50 A. 3HCl Al(OH)3 → AlCl HOH B. HCl KOH → KCl + HOH C. HCN KOH → KCN + HOH
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5. Match the Curve with the Reaction
pH Volume 1.0 M NH3 added 14 7 30 60 A. 3HCl Al(OH)3 → AlCl HOH B. HCl KOH → KCl + HOH C. HCN KOH → KCN + HOH
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6. Match the Curve with the Reaction
pH Volume 1.0 M NH3 added 14 7 30 60 A. 3HCl Al(OH)3 → AlCl HOH B. HCl KOH → KCl + HOH C. HCN KOH → KCN + HOH
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Features of the Strong Acid-Strong Base Titration Curve
The pH starts out low, reflecting the high [H3O+] of the strong acid and increases gradually as acid is neutralized by the added base. Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point. Beyond this steep portion, the pH increases slowly as more base is added.
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HPr = Propionic Acid
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The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve The initial pH is higher. A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point. The pH at the equivalence point is greater than 7.00. The steep rise interval is less pronounced.
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