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Chem. 1 Notes: Unit 13 Acids, Bases & Aqueous Equilibrium

Chem. 1 Notes: Unit 13 Acids, Bases & Aqueous Equilibrium.

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Chem. 1 Notes: Unit 13 Acids, Bases & Aqueous Equilibrium

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  1. Chem. 1 Notes: Unit 13Acids, Bases & Aqueous Equilibrium

  2. I. General Properties: ACIDBASESour BitterExamples of common Examples of bases;acidic substances; ammonia, baking vinegar (acetic acid) soda, oven cleaner citric acid, battery acid milk(sulfuric acid) Acid + metal ___> H2 Acids feel slippery when wet Acid + Base ___> Salt + Water Both change color of indicator dyes

  3. Examples of acid/base reactions;NaOH + HCl H2O + NaClBase acid water saltH2SO4 + 2CaOH 2H2O + CaSO4acid base water salt

  4. II. DefinitionsA. Arrhenius (traditional) 1. Acids – dissociate(break apart) in water to produce H+1 (hydrogen ions)/ H3O+1 (hydronium ions) ex. HCl___> H+1 + Cl-1 (H+1is very reactive so…)HCl+ H2O ___> H3O+1 + Cl-1 (H3O+1 = hydronium ion) 2. Bases –dissociate in water to produce OH-1(hydroxide ions)ex. NaOH___> Na+1 + OH-1

  5. 3. Reactions - Acid + Base ___> Salt + Waterex. HCl(aq) + NaOH(aq)___> NaCl(aq) + H2O(l) (molecular eqn.)H+1(aq) + Cl-1(aq)+ Na+1(aq) + OH-1(aq)___> Na+1(aq) + Cl-1(aq) + H2O(l) (ionic eqn.)H+1(aq) + OH-1(aq)___> H2O(l)“spectator” ions removed(net ionic eqn.)

  6. B. Bronsted-Lowry acids and bases1. Acids – proton (H+) donors2. Bases – proton (H+) acceptors3. Reactions - Acid + Base ___> Conjugate acid + Conjugate base(acid donates proton to a base and becomes a conjugate base; base accepts a proton and becomes a conjugate acid) ex. HCl + H2O ___> H3O++ Cl- acid base c. acid c. base ex. NH3 + H2O ___> NH4++ OH- base acid c. acid c. base * some substances (water) can act as an acid or a base = amphoteric I would know what this word means!!!

  7. III. Acid NomenclatureA. Terms1. monoprotic acids – have one proton (H+) to donate ex. HCl___> H++ Cl-2. polyprotic acids – have more than one proton (H+) to donateex. H3PO4___> H++ H2PO4- (first proton)H2PO4-1___> H++ HPO4-2 (second proton) HPO4-2___> H++ PO4-3 (third proton)

  8. B. Binary acids – H+ + monatomic anion ex. HCl Names = “hydro-“ + root of monatomic anion + “-ic”HCl = hydrochloric acid HI = hydroiodic acidC. Ternary acids – H+1 + polyatomic ion ex. HNO3If polyatomic ions suffix = “-ate” name = root of polyatomic ion + “-ic + acid Ex. HNO3 = (H+1 + nitrate ion) = nitric acidIf polyatomic ions suffix = “-ite” name = root of polyatomic ion + “-ous + acid Ex. HNO2 = (H+1 + nitrite ion) = nitrous acid

  9. IV. Measurement of [Acid] and [Base] A. pH scale ( “power of hydrogen”) – usually very small #’s, so a logarithmic scale is used – measured by indicator dyes when a solution reaches a certain pH, it turns a specific color to indicate that pH 0-6.9 = acidic 7.0 = neutral 7.1 – 14 = basic

  10. B. Equations (yes, you WILL need to know these)1. pH = -log [H3O+1] the [ ] mean concentration of the H+ or H3O+1 ions2. [H3O+1] = 10-pH3. pOH = -log [OH-1]4. [OH-1] = 10-pOH5. pH + pOH = 14.06. [H3O+1] x [OH-1] = 1.0 x 10-14

  11. V. Dissociation Constants -we are now talking about the known rate at which an acid or base breaks apart(dissociates) in solution A. Water: H2O(l) + H2O(l)___> H3O+1(aq) + OH-1(aq) K = [H3O+1] x [OH-1] [H2O]2 * [H2O] is constant so K x [H2O] = Kw = [H3O+1] x [OH-1]Kw = (1.0 x 10-7) x (1.0 x 10-7) at 25oCKw = 1.0 x 10-14 * amounts of H3O+1 and OH-1 determine pH if [H3O+1] > [OH-1] = acidic (pH < 7.0) if [H3O+1] = [OH-1] = neutral (pH = 7.0) if [H3O+1] < [OH-1] = basic (pH > 7.0)

  12. B. Acids1. Strong – dissociate 100% -it all breaks apart HA + H2O _100%__> H3O+1 + A-1 [I] 0.5 0.0 0.0 [C] -0.5 +0.5 +0.5(* 100% of HA dissociates) [E] 0.0 0.5 0.5(* use [H3O+1] to find pH)acid dissociation constant, Ka = [H3O+1] x [A-1] = (0.5) x (0.5) = very large [HA] (0.0) strong acids – 3 binary = HCl, HBr, HI 4 ternary = H2SO4, HNO3, HClO3, HClO4

  13. 2. Weak – dissociate < 100% HA + H2O _______> H3O+1 + A-1 [I] 0.5 0.0 0.0 [C] 0.5 - x +x +x (* <100% of HA dissociates) [E] 0.5 – x xx (* use [H3O+1] to find pH)acid dissociation constant, Ka = [H3O+1] x [A-1] = x2 [HA] [HA] = small

  14. A. Finding [equilib] and pH from Ka and [initial]1. Write a balanced equation for dissociation rxn. ICE table2. Use Ka and [E] to find value of “x” (H3O+1) Ex. Find [equilib.] for all species and pH for a solution of 0.1M HCN in H2O. HCN + H2O ___> H3O+1 + CN-1[I] 0.1 M 0.0 M 0.0 M[C] -x +x +x [E] 0.1 – x xx3. Ka = 4.9 x 10-10 = [H3O+1] x [CN-1] = x2 [HCN] (0.1 – x)

  15. (*since Ka is small compared to initial [HCN], assume D[HCN] (x) = 0) Ka = 4.9 x 10-10 = x2 0.1x = 7.0 x 10-6 so [HCN] = 0.1 - 7.0 x 10-6 = 0.1 M [H3O+1] = 7.0 x 10-6 [CN-1] = 7.0 x 10-6[OH-1] = Kw/[H3O+1] = 1.0 x 10-14= 1.4 x 10-97.0 x 10-6 pH = - log [H3O+1] = - log (7.0 x 10-6) = 5.15pOH = 14.00 – 5.15 = 8.85

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