1 / 15

Acid & Bases

Acid & Bases. Arrhenius: Acids are proton (H + ) sources and bases are hydroxide ion (OH - ) sources. E.g. HCl is an acid and NaOH a base

gwenllian-baker
Download Presentation

Acid & Bases

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. Acid & Bases • Arrhenius: Acids are proton (H+) sources andbases are hydroxide ion (OH-) sources. E.g. HCl is an acid and NaOH a base • Brønsted-Lowry: Acids are proton sources and bases are proton acceptors. E.g. HCl is an acid and NH3 a base, these form the conjugate base Cl- & NH4+. • Lewis: Acids are electron pair acceptors and bases are electron pair donors. E.g. AlCl3 & :NH3. to form Cl3Al:NH3

  2. Acid & BasesAcid Strength Acid strength describes the position of the equilibrium or the size of Keq HX  H+ + X-, KA= [H+][X-]/[HX] for example: H2SO4 H+ + HSO4- KA ~ 1.0*103 HSO4- H+ + SO4- KA ~ 1.2*10-2 In the first case the reaction essentially goes to completion, in the second HSO4- hardly dissociates.

  3. Acid-Base Equilibria • For now, an acid is a proton (H+) donor and a base is either a proton acceptor or a hydroxide (OH-) donor. Water will be the solvent throughout this discussion. • Water dissociates to give both a proton and a hydroxide ion. This may be written several ways. I choose to write it in the simplest, least correct, way -- H2O  H++ OH-

  4. Acid-Base Equilibria • this equilibrium may be (poorly) modeled by the equation: Keq= [H+][OH-]/[H2O] where Keq = 1.0*10-14 at 25 oC & the activity of water is 1. • One may rewrite the equation as: Keq*[H2O]= KW = [H+][OH-] = 1.0*10-14. Also, in neutral solution, [H+]=[OH-]= (1.0*10-14)1/2 = 1.0*10-7

  5. Acid-Base Equilibria • solving for [H+], we get: [H+]=[OH-]= (KW)1/2 • Or, taking the Log10 of both sides we get: Log10KW = Log10[H+] + Log10[OH-] = -14 • multiplying both sides by -1 gives • -Log10KW = -Log10[H+] + -Log10[OH-] = 14

  6. Acid-Base Equilibria • defining “p” to mean "1/Log10 " or -Log10 , we can rewrite our equation to become: pKW = pH + pOH = 14 • Thus, for water or a "neutral" solution at 25oC, the pH = pOH = 14/2 = 7 or both the [H+] and [OH-]= 10-7M

  7. Acid-Base Equilibria • For any other type solution, the hydrogen or hydroxide ion concentrations will depend on BOTH the dissociation of water and ions contributed by other components of a solution. For example, if we make a 0.20M solution of nitric acid the hydrogen ion concentration would depend on the hydrogen ion from the nitric acid and from the dissociation of water. Let’s model this on the blackboard.

  8. Acid-Base Equilibria • A MAJOR PITFALL for the lazy or unwary student is to ignore ion sources. This person would say, "we need consider only the strong acid or strong base concentration and may always ignore the contribution from water."NOT! • If you remember that the ion concentrations in equilibrium expressions are ALWAYS TOTALS, Acid-base equilibria, indeed all solution equilibria, are much easier to understand.

  9. Acid-Base Equilibria • Let’s look at what happens if we have a weak acid instead of a strong one. HA  H+ + A- KA = [H+]T[A-] [HA] where the “T” means total hydrogen

  10. Acid-Base Equilibria • Solving for [H+]T, we have: [H+]T= [Ac-] + [OH-] [H+]T = [Ac-] + Kw/[H+] [Ac-] = [H+]T - Kw/[H+] & KA = [H+][Ac-]/[HAc] = [H+] ([H+] - Kw/[H+] ) CA - ([H+] - Kw/[H+] ) = ([H+]2 + Kw CA – {([H+]2 – Kw)/[H+]}

  11. Acid-Base Equilibria If [H+]T > ~ 10-5, this simplifies to: [H+]T = KA*[HAc]/[Ac-] Or, in this case, the water contribution is negligible compared to the proton from the acetic acid.

  12. Acid-Base Equilibria • Let's look at what's left: [H+]T = KA*[HA]/[A-] CAM 0 0 HA  H+ + A- & KA = [H+][A-]/[HA] (CA-x)M xM xM or KA = X*X/(CA-X), expanding we find X2 + CAX - KA CA = 0, a quadratic or, if X is much smaller than CA, KA= X2/CA, & X=[H+]= (CAKA)0.5

  13. Acid-Base Equilibria • If we have a very dilute SA solution: HAH+ + A- & [H+] = [A-] + [OH-] [H+] = [A-] + KW/ [H+] ; [A-] = CA solving for CA; CA = ([H+]2 – Kw)/[H+] or [H+]2 – CA[H+] - Kw = 0 For example, if CA = 1.0*10-7, pH ~ 6.80

  14. Acid-Base Equilibria Alternatively: If one adds enough nitric acid to water to prepare a 1*10-7M solution, the only equilibrium is: H2O  H++ OH- KW = [H+][OH-]

  15. Digression – base 10 Logs 4=2*2, log 4=.30+.30 = .60 2.5 = 25*10-1= 5*5*10-1 log 2.5= .70+.70+(-1) = .40 Antilog of – 0.60 = antilog(-1)+.40 = 2.5*10-1 or 0.25 if you multiply #’s, logs are added divide #’s logs are subtracted take a # to a power, multiply log times power

More Related