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UH BTEC 1322 - Titration
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CHEM 1332 1st Edition Lecture 18 Outline of Last Lecture - Molarity- Spectator ions- Neutralization- Solving pH equationsOutline of Current Lecture - Four steps- Titration- ExamplesCurrent Lecture- The four steps to solving pH problems are:- First make sure you have the right concentration- Second delete spectator ions- Third is neutralization- Fourth is the calculations identity what in the solution affects or leads to pH- Titration is taking an A and adding a little B - There are two main reasons why this is done first you might not know much about B so you want to know exactly when you have added enough .The second reason is to help you study the reaction as it is happening - The reactions that we will first study is acids and bases and you take your acid or base and add to it base or acid bit by bit it is important to measure the pH- The titration curve for this type of reaction is pH versus volume added- An example for a strong acid and strong base is 20 mL of 0.1 M HCl add 0.1 M NaOH- Find the pH when 0 mL of NaOH added which is 0.1 M HCl which the pH=-log(0.1)=1- The pH of 5mL NaOH added - Step one make sure you have the right concentration:- New HCl concentration is (0.1M)(20mL)/25mL=0.08 M- New NaOH concentration is (0.1M)(5mL)/25mL=0.02 M- Step two delete spectator ions 0.08 HCl=0.08H 0.02NaOH=0.02 OH+¿H¿−¿OH¿These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.S 0.08 0.02C - 0.02 - 0.02An 0.06 0- Left with 0.06 M +¿H¿- pH=-log(0.06)=1.2- 10 mL of NaOH solution - New HCl=(0.1)(20)/30=0.06666- New NaOH=(0.1)(10)/30=0.03333+¿H¿−¿OH¿S 0.06666 0.03333C - 0.03333 - 0.03333An 0.03333 0- pH=-log(0.03333)=1.5- 19.9 mL of 0.1 M NaOH to 20 mL of 0.1 M HCl- New HCl=(0.1)(20)/39.9=0.0501- New NaOH=(0.1)(19.9)/39.9=0.0499+¿H¿−¿OH¿S 0.0501 0.0499C - 0.0499 - 0.0499An 0.0002 0- pH=-log(0.0002)=3.7- 20 mL of 0.1 NaOH to 20 mL of 0.1 HCl- New HCl=(0.1)(20)/40=0.05- New NaOH=(0.1)(20)/40=0.05 (this is the equivalent point. The pH is equal to 7+¿H¿−¿OH¿S 0.05 0.05C - 0.05 - 0.05An 0 0- 20.1 mL of 01. M NaOH added to 20 mL of 0.1 M HCl- New HCl=(0.1)(20)/40.1=0.0499- New NaOH=(0.1)(20.1)/40.1=0.05+¿H¿−¿OH¿S 0.0499 0.0500C - 0.0499 - 0.0499An 0 0.002- pOH=-log(0.002)=3.7 14-3.7= 10.3- Weak acid + strong base- 20 mL of 0.1 M C H3COOH Ka=1.8 ×10−5- Add 0.2 M KOH (aq)- Add 0 mL KOH- 20 mL of 0.1 M C H3COOH pH=2.87- Add 3 mL KOH- New C H3COOH=(0.1)(20)/40=0.0870 M- New KOH=(0.2)(3)/23=0.0261 MC H3COOH−¿OH¿C H3COOS 0.0870 0.0261 0C -0.0261 -0.0261 +0.0261An 0.0609 0 0.0261- They are with 10:1 or 1:10 ratio so it is a buffer solutionC H3COOH+ H2O →H3OC H3COOS 0.0609 0 0.0261C -X X +XE 0.0609-X X 0.0261 +X- KA(C H3COOH ¿=1.8× 10−5=(X )(0.0261+ X )(0.0609−X ) Ka is small so you can assume one X is insignificant solve get X=4.2×10−5 pH=-log(4.2×10−5¿=4.38Key TermsTitration- lab method to determine the unknown concentration of an identified or unknown


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