1 / 53

Acids, Acids Everywhere

A cids and or systems that are controlled by acidity are in almost every aspect of our day to day lives. From: O ur morning coffee…(a base) Digestion of food in our stomachs. To: The production of soap and shampoo The gas we put in our cars

neila
Download Presentation

Acids, Acids Everywhere

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Acids and or systems that are controlled by acidity are in almost every aspect of our day to day lives. From: Our morning coffee…(a base) Digestion of food in our stomachs. To: The production of soap and shampoo The gas we put in our cars Pretty much everything is affected/created by acids and bases. Acids, Acids Everywhere

  2. In our previous work with acids and bases we learned that certain acids are stronger than others. 6 Strong Acids?: So why are certain acids strong and other are weaker??? Names? HCl HBr HI HNO3 H2SO4 HClO4 Hydro Chloric Acid Hydro Bromic Acid Hydro Iodic Acid Nitric Acid Sulfuric Acid Perchloric Acid Equilibrium and Acid/Base Strength

  3. Previously we have learned that in order to produce and acid or basic solution chemicals have to react with water to produce?(ions)? H3O+ Hydronium ACIDS OH- Hydroxide BASES Now that we have looked at the equilibrium that exists between reactants and Products we can use this knowledge with acids/bases. H30+, OH- H20

  4. Even pure water has an equilibrium between the liquid (H2O) and ion(H3O+/OH-) forms. Because of this equilibrium, even pure water has some H+ and OH- ions in it. The amount of ions present (%ionization) in the equilibrium of water is very small, something like 2 out of 1,000,000 molecules. The % ionization (the chance ions are created) is so small that the corresponding Kc is also very small. Because water is involved in every solution on earth its very important and has a special symbol and it equilibrium constant is very important.

  5. If you write the Kc expression for the ionization of water you notice something familiar. 2 H2O(l) H3O+(aq) + OH-(aq) [H3O+(aq)][OH-(aq)] [(H2O(l)]2 No liquids!

  6. Whenever you deal with pH and pOH, all you need is!..... ??? Perfect Diamond Shape pH & pOH…..again

  7. Practice

  8. We’ve seen previously that a 1.0mol/L concentration of an acid can have a pH anywhere from -1 to 6.9 So it’s the same concentration why does and its an acid why doesn’t it have the same pH? ***The concentration of the acid has nothing to do with how strong it is. (its pH…how many H3O+(aq) ions it has).*** If an acids strength is determined by how many H3O+(aq) then really acid strength is determined by the equilibrium between: Acid-H(aq) + H2O(l) H3O+(aq) %?

  9. Question HCl(aq) It’s a strong acid so its percent reaction is 100%. Given Information ? 1:1 Ratio 100% C=0.10 mol/L C=0.10 mol/L C=? C=0.1 mol/L Just saying….. Keep that in mind Hint Hint… Calculating Concentration of H3O+(aq) Using % Equilibrium [H3O+(aq)]= %Reaction X [HCl(aq)] = (1) X 0.10 mol/L = 0.10 mol/L

  10. % Reaction must be in a decimal! [H3O+(aq)]= %Reaction X[CH3COOH(aq)] = (0.013) X 0.10 mol/L = 0.0013 mol/L ~ 1.3 x10-3mol/L Practice

  11. As with all formulas, you can also re-arrange the % reaction formula to solve for any of the three variables. [H3O+(aq)]= %reaction X [Acid-H(aq)] [Acid-H(aq)] [Acid-H(aq)] Re-areanging the % Reaction Formula [H3O+(aq)]= %reaction [Acid-H(aq)] (100) X

  12. But what about compounds that can accept AND donate H+(aq)? H2CO3(aq) Sodium BiCarbonate H2CO3(aq) + H2O(l)HCO3-(aq)+ H3O+(aq) HCO3-(aq) + H2O(l) H2CO32-(aq) + OH-(aq) HCO3-(aq) + H2O(l)  CO32-(aq) + H3O+(aq) So How do we know what its going to do? Acid or Base? Base Acid

  13. Acids: Donate Protons (H+) and turn H2O into H3O+ ions. Bases: Accepts Protons turning H2O molecules into OH- ions. Johannes Bronstead (1879-1947 Thomas Lowry (1874-1936 The Proton Transfer ConceptTheory For Acids and Bases(Bronsted-Lowry Definition of Acids and Bases) Amphoteric: Is an entity that can act as an acid (donate protons) or a base (accepting protons). Amphiprotic: Is a entity that can donate/accept ONE SINGLE proton.

  14. Little Extra: The H2O accepts the proton…so by the definition it is a what? In a chemical equation where an acid donates a proton their MUSTALSO be a BASE to accept it….ALWAYS. Bronsted-Lowry ACIDSProton Donors

  15. Little Extra: Again, if H20 donates the proton then NH3 must accept it….making the NH3 a what??? And yet again, there are ALWAYS an acid AND a base. Bronsted-Lowry BASESProton Acceptors

  16. So if an acid donates an H+ and a base accepts an H+(aq)……does a base turn into an acid after it accepts an H+(aq )??? And vice versa when an acid donates its proton?? Labeling Bronsted-Lowry Practice

  17. When an acid reacts in solution it donates its H+(aq) to the H2O(l) to create H3O+(aq). + H1+(aq)  H3O+(aq) H2O(l) But what happens to the other half of the acid after the H is gone? Conjugate Base Base Conjugate Acids and Bases H3O++Cl-(aq) + H20(l) HCl(aq) Conjugate Acid Acid

  18. Acetic acid (vinegar) is mixed with water and ionizes to produce hydronium ions. But because acetic acid is such a weak acid only a small % of the acid ionizes to produce hydronium ions in the following equation. Label the different chemicals in the equation as; acid, conjugate base, base, conjugate acid. Conjugate Acid Acid Base Identifying Conjugate Acids/Bases Conjugate Base

  19. In general acids contain H’s (at least one) in their chemical formula’s. When acids react in a chemical reaction they donate the H+(aq) to either: Water to form H3O+(aq). Or a base to neutralize OH-(aq) molecules, which forms……H2O(l). Acid General Form: Use to be neutral. Lost H+(aq). Leftover is 0-(+1)= -1 Acid & Base Conjugates General Form

  20. So the big question now is; When you mix a bunch of chemicals together, a lot of chemicals can act as either an acid (accepting) or a base….how do we know which chemicals are going to do what? Lots of different chemicals can act as a base, and lots of chemicals can act as an acid. How do we know which chemicals will act like an acid and which will act like a base? Survival of the fittest…..err strongest. Predicting Acid-Base ReactionsAcid/Base Strength

  21. So we know there are six STRONG acids….so they win. What about the rest that aren’t “Strong”? Pages 8 and 9 In your data booklet rates the strengths of all acids and bases. Don’t memorize, just familiarize…. AKA, you will always have the data booklet to help you remember, but!***Memorizing the most common ones we use will be helpful if you do any chemistry later in life*** Predicting Acid-Base ReactionsAcid/Base Strength

  22. ST R ONG AcIDS HTGNERTS GNISAERCED DECREASING STRENGTH Strong Bases ANYTHING WITH OH (hydroxide) IN IT IS A STRONG BASE!

  23. 1. Use solubility rules to list all entities present. Sodium Hydroxide (NaOH) is mixed with Vinegar (CH3COOH). Solubility Table Its Ionic Its Molecular H2O(l) NaOH CH3COOH (?) (aq) (aq) (?) It’s a weak Acid H2O(l) CH3COOH(aq) OH-(aq) Na+(aq) 5 Steps To Predicting Acid/Base Reaction Equations

  24. 2. Use the acid/base strength table to label each entity as acid (A) or base (B). A A B B Acid/Base Table H2O(l) CH3COOH(aq) OH-(aq) Na+(aq) 5 Steps To Predicting Acid/Base Reaction Equations

  25. 3. Use the acid/base strength table to label the strongest acid (SA) and strongest base (SB). A SA A H2O(l) CH3COOH(aq) OH-(aq) Na+(aq) B SB B 5 Steps To Predicting Acid/Base Reaction Equations

  26. 4. Write a balanced chemical equation with conjugate base (acids ½) and conjugate acid (base ½). Also showing the proton transfer. SA + SB SA(conjugate)+ SB(conjugate) A SA H2O(l) CH3COOH(aq) OH-(aq) Na+(aq) B SB H+ 5 Steps To Predicting Acid/Base Reaction Equations H2O(l) CH3COOH(aq) OH-(aq)  CH3COO-(aq) + +

  27. 5. Predict the position of the equilibrium….write % of reaction over the reaction arrow. To Predict the % reaction you use the following generalisation: Acid/Base Table H+ 5 Steps To Predicting Acid/Base Reaction Equations H2O(l) CH3COOH(aq) OH-(aq)  CH3COO-(aq) + +

  28. 5. Predict the position of the equilibrium….write % of reaction over the reaction arrow. To Predict the % reaction you use the following generalisation: >50% H2O(l) CH3COOH(aq) OH-(aq)  CH3COO-(aq) + + DONE! FAIT! אני עושה! Я сделал! 5 Steps To Predicting Acid/Base Reaction Equations

  29. First Step is Denial. Second Step is Anger. The third step is Bargaining. The fourth step is Depression. The fifth step is Acceptance. These are the 5 steps to overcoming grief. These have absolutely nothing to do with Chemistry. A SummaryThe 5 Step Program…Like A.A Who wrote them down? 

  30. A SummaryThe 5 Step Program…Like A.A Paraphrase it….DONT WRITE IT ALL DOWN WORD FOR WORD!!!

  31. When you talk about acid and base strength you show it just like you did with Kc. But! Instead of Kc; Ka for Acid Kb for Bases The formula is still the same. PRODCUTS/REACTANTS. [PRODUCTS] [REACTANTS]

  32. Ka and Kb equations still work the same…. You still set them up the same and use the balancing coefficients as exponents. You can still put in the information you know from the question and solve for what you don’t know. Just rearrange it. Ka and Kb Equations

  33. Well, its an acid so Ka. Ka=[PRODUCTS] [REACTANTS] The Mission: Write an equilibriumn expression for hydroflouric acid reacting in water to produce hydronium and HF’s conjugate base. Done. Right? =[H3O+(aq)][F-(aq)] [HF(aq)][H20(l)] Practice!

  34. This is just more practice of what we already learned earlier. Because K a and Kb are the same as Kc, using an I.C.E table is just doing the same as before. REMEMBER! Whenever your given an equilibrium question, Kc, Ka, Kb, Kwhatever….you need a balanced equation! Ka and Kb AND I.C.E Tables.

  35. Wants Ka…what’s the first thing we need to do? Ka=[PRODUCTS] [REACTANTS] =[H3O+(aq)][CH3COO-(aq)] [CH3COOH(aq)]][H20(l)] So now just substitute what you know from the question into the equation you just made. It’s just that easy…….right….. Practice!Easy…..(5/10)

  36. ???? =[H3O+(aq)][CH3COO-(aq)] [1.00] What about CH3COO-??? Well since we have a balanced chemical equation we can calculate CH3COO-...(1/1) Do We know anything else??? [H3O+(aq)]= 10-pH [H3O+(aq)]= 10-2.38 [H3O+(aq)]= 0.0042 mol/L =[H3O+(aq)][CH3COO-(aq)] [CH3COOH(aq)] Practice! =[0.0042][CH3COO-(aq)] [1.00] =[0.0042][0.0042] [1.00] =0.000017 ~ 1.7 x 10-5

  37. Balanced Chemical Equation. What’s the first thing we want to do/make/write down? Hydrocyanic Acid(aq) + H2O(l) H3O+(aq)+ CN-(aq) Hydrocyanic Acid –H(aq) HCN(aq) How can we find this chemical formula? How can we find this chemical formula? Practice!Hard-ish (8/10)

  38. Balanced Chemical Equation. + H2O(l) H3O+(aq)+ CN-(aq) HCN(aq) Next? =[H3O+(aq)][CN-(aq)] [HCN(aq)]][H20(l)] Ka=[PRODUCTS] [REACTANTS] =[H3O+(aq)][CN-(aq)] [HCN(aq)] Write a Ka expression for the reaction. Well, that’s not a lot to go by. But because we have more than one unknown, it’s a sign to use an I.C.E table. 6.2 X 10-10=[?][?] [0.500] Ka=[H3O+(aq)][CN-(aq)] [HCN(aq)] 6.2 X 10-10=[H3O+(aq)][CN-(aq)] [HCN(aq)] 6.2 X 10-10=[?][CN-(aq)] [HCN(aq)] 6.2 X 10-10=[?][?] [HCN(aq)] Practice! ???? ???? ???? ????

  39. 6.2 X 10-10=[?][?] [0.500] 6.2 X 10-10=[x][x] [0.500-x] 6.2 X 10-10=[x][x] [0.500] So…. + H2O(l) H3O+(aq)+ CN-(aq) HCN(aq) *** Because the Ka is so small….6.2 x10-10 =0.00000000062 WE ASSUME 0.500-x = 0.500 still Practice! What about the change? What do we know from the question? Are they going to increase or decrease as the reaction occurs? Does the question tell us anything about H3O+ or CN-??

  40. Now that we have two unknowns and their BOTH X (because their the same amounts…..because of the 1:1 ratio that we got from the balanced chemical equation. Re-arrange and Solve for X. 6.2 X 10-10=[x][x] [0.500] 6.2 X 10-10= X2__ [0.500] (500) (500) 0.00000031= X2 X2 = 0.00000000031 Practice! X = 1.760681686 x 10-5 ~ 1.76 x 10-5

  41. ~ 1.76 x 10-5 [H3O+(aq)]=1.76 x 10-5 pH = -log[H3O+(aq)] pH = -log(1.76 x 10-5) pH = 4.754319153 ~ 4.75 Practice!

  42. A little while back we learned what Kw is... Anybody remember what it is? The value of it? So, Kw is the water ionization CONSTANT, so that means its always 1.0 x 10-14. 1.0 x 10-14 might look familiar because we also learned about it with pH and pOH. Water Ionization ConstantKw…Again…

  43. But now that we know more about equilibrium constants and acids and bases we can change the formula a bit more. But now that we know that Ka is related to [H+(aq)]/[H30+(aq)] and Kb is related to we can make a NEW formula! This is really useful because if you ever know ka or kb, you can find the other and vice versa. Write it down! Water Ionization ConstantKw…Again… Like all formulas, this one can be re-arrange to solve for what you don’t know. Kw =KaXKb

  44. As always, the first step is: Balanced Chemical Equation Ka Kb So where would we look for find the Kb? But the data booklet “Relative Strengths of Acids and Bases” table on has Ka’s… So use the water ionization formula. Put in what you know. Solve for the one you don’t Kb in this case. Kw =KaXKb Ka/Kb Practice 1.0 x 10-14 =5.6 x 10-10XKb = Kb1.0 x 10-14 5.6 x 10-10 1.785714286 x10-5 ~ 1.8 x 10-5

  45. Big things to remember about titration: Titrant Standard Solution End Point  Equivalence Point  The stuff being dripped out of the burette. The solution in the beaker that you KNOW the concentration of. Is the point, the exact drop that causes the indicator in the standard solution to change color. THEORETICAL point at which the amount of titrant moles = Standard solution moles.

  46. Buffering Region • “The Waterfall” or “Change Region” • Buffering Region • Not a big change in pH • During this time OH and HCl are reacting (canceling each other out) making H2O • At this point the last moles of NaOH are being neutralized by added moles of HCl. pH Curve’s Unique ShapeBreaking It Down Into Regions • After the last moles of NaOH are gone, a single drop of HCl makes a huge change in pH. pH curve for Strong Acid (HCl) and Strong Base (NaOH)

  47. 3. Buffering Region • All NaOH is gone, all that’s left is acid. • The pH levels off at the pH of the pure acid. • HCl 0.8-1.7 • CH3COOH  2.4-3.4 • H2SO4 1.1-1.9 • The pH range depends on how concentrated the acid is pH Curve’s Unique ShapeBreaking It Down Into Regions pH curve for Strong Acid (HCl) and Strong Base (NaOH)

More Related