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ACID-BASE

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ACID-BASE

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  1. ACID-BASE EQUILIBRIA

  2. TOPICS: ACID DISSOCIATION CONSTANT BASE DISSOCIATION CONSTANT IONIC PRODUCT OF WATER

  3. BACKGROUND: ACID: A strong acid is defined as an acid that ionizes completely in aqueous solutions producing a high concentration of hydrogen ions. BASE: Bases are also classified as strong and weak bases. Bases that dissolve in water are referred to as alkali. strong bases ionize completely in water whereas weak bases do not ionize completely in aqueous solutions.

  4. ACID DISSOCIATION CONSTANT (ka) A weak acid can be represented by the formula HA and dissociates according to the equilibrium: HA + H2O H3O+ + A- HA(aq) H+(aq) + A-(aq)

  5. ACID DISSOCIATION CONSTANT (Ka) A weak acid dissociates to a small extent in aqueous solution until equilibrium is achieved. The equilibrium expression written for the reaction is called the acid dissociation constant (Ka) The acid dissociation constant (Ka) can be used to compare the strengths of acids. Ka= [H3O+] [A-] Ka= [H+] [A-] [HA] [HA]

  6. ACID DISSOCIATION CONT’D E.g. - Reaction in which the weak acid, ethanoic acid dissolves in water. CH3COOH (aq)CH3COO-(aq) +H+(aq) Ka = [CH3COO-(aq)] [H+(aq)] [CH3COOH (aq)] Since water is the solvent, it is in large excess and is excluded from the equation. Water is only included in an equilibrium expression if it is a reactant or a product in a chemical reaction.

  7. NOTE: If the concentration of the acid is known together with the concentration of the ions, a value of ka can be easily calculated. A large ka means that the acid is highly dissociated therefore it is a strong acid.

  8. pka pka is defined as the negative of the log to base 10 of ka. pka = -lg ka NB:- pka is not the pH of a weak acid. pka can be USED to calculate the pH of a weak acid.

  9. CALCULATING THE pH OF A WEAK ACID Q: Calculate the pH of a solution of acetic acid of concentration 0.10 mol dm-3, given that the ka of acetic acid is 1.7 * 10-5. A: The equation for the reaction is: HC2H3O2 (aq) H+(aq) + C2H3O2-(aq)\ Assuming that x moles of the acid had dissociated to produce x moles of the C2H3O2- ions and x moles of H+ ions: ka= [C2H3O2-] [H+] = (X)(X) [HC2H3O2] (0.1-X)

  10. The solution of quadratic equations is not required by CAPE so we’re going to forget about the denominator of the fraction. i.e. ka= x2 0.1 Given that ka = 1.7 * 10-5, 1.7 * 10-5 = x2 (1.7 * 10-5)(0.1)= x2 0.1 1.7 * 10-6 = x2 √ 1.7 * 10-6 = x Therefore x= 1.3 * 10-3 X= [H+] = [C3H3O2-] = 1.3 * 10-3 Solving for pH, we use the pH equation: pH= -lg [H+] = -lg [1.3 * 10-3] = 2.9

  11. BASE DISSOCIATION CONSTANT (kb) Base dissociation constant can be used to compare the strengths of bases. The base dissociation constant can also be used to calculate the concentrations of ions. Just as acids, an equilibrium expression can also be written for the dissociation of a base: B (aq) + H2O (l) BH+(aq) + OH-(aq) kb = [BH+(aq) ] [OH-(aq) ] [B (aq) ]

  12. Eg:Consider the reaction of methylamine with water. CH3 NH2(aq) + H2O (l) CH3NH+3(aq) + OH-(aq) kb = [CH3NH+3] [OH-] [CH3NH2] Taking negative logarithm to base 10 gives: pkb = -lg (kb) NOTE: A small kb value means that the base is dissociated to a small extent therefore it is a weak base. The stronger the base the larger the kb value and hence the smaller the pkb value.

  13. IONIC PRODUCT OF WATER A sample of pure water will contain a small quantity of ions (H+ and OH-) produced from the self-ionization of water. Water at 250C dissociates to a very small extent producing a minute quantity of the hydrogen cation and hydroxide anion. An expression can be written for the dissociation as dynamic equilibrium is achieved. H2O (l) H+(aq) + OH-(aq) kc = [H+] [OH-] kc [H2O] = [H+] [OH-} [H2O]

  14. kc is a constant and the concentration of water is considered to be a constant because so little of the water has dissociated that the change is negligible. Two constants when multiplied together give a new constant, which we call kw, the ionic product of water. At 250C it has been found that the concentration of H+ and OH- is 1*10-7 mol dm-3 respectively. kw = [H+] [OH-] = (10-7 mol dm-3) *(10-7 mol dm-3) = 10-14 mol2 dm-6

  15. THE RELATIONSHIP BETWEEN Ka AND Kb The relationship between Ka of the conjugate acid and Kb of the base is as follows: Consider the acetic acid equilibrium: HC2H3O2(aq) H+(aq) + C2H3O2-(aq) Ka=1.7*10-5 Next consider then hydrolysis of water by acetic acid’s conjugate, the acetate ion: C2H3O2-(aq) + H2O (l) HC2H3O2(aq) + OH-(aq) What if you wanted to calculate the Kb for this equilibrium?

  16. There is a relationship between these two expressions that may not have been obvious when you looked at them in this way. HC2H3O2(aq) H+(aq) + C2H3O2-(aq) C2H3O2- (aq) + H20 (l) HC2H3O2(aq) + OH-(aq) If you add the two equations together, you will find that the net equation is : H2O (l) H+(aq) + OH-(aq)

  17. This is the equation for the self-ionisation of water, from which Kw was derived. The first is an equation that would use to calculate the Ka of acetic acid. The second is the equation used to calculate Kb for the conjugate base. The third, which is derived from these two, is the formula for calculating Kw. The three constants are all related in the equation shown below. Kw = KaKb