Acid Base Equations General Form of Dissociation Acid HA H A or H 2 O A H AOH Base BOH B OH or H 2 O B ID: 1011467
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1. Acids and BasesChapter 14
2. Acid Base Equations
3. General Form of DissociationAcidHA H+ + A- orH2O + A H+ + AOH-BaseBOH B+ + OH- or H2O + B BH+ + OH-Strong acids/bases will have a , weak acids/bases will have a ⇌
4. Some weak acids and basesAcids BasesCarbonic acid CarbonateAcetic acid AcetateAmmonium Ammonia triethylamine(CH3CH2)3NH+ (CH3CH2)3N
5. Strong acidsAcidformulaAcidFormulaHydrochloric acidHClSulfuric AcidH2SO4Hydrobromic acidHBrNitric AcidHNO3Hydriodic acidHIPerchloric AcidHClO4Chloric AcidHClO3
6. Strong BasesNameFormulaNameFormulaSodium HydroxideNaOHCalcium HydroxideCa(OH)2Potassium HydroxideKOHStrontium HydroxideSr(OH)2All group 1 metalsBarium HydroxideBa(OH)2these make a lightning bolton the periodic table!
7. Strong acids and basesStrong acids and bases are not at equilibrium, there is no reverse reaction.Strong acids and bases will never be formed in a net ionic equation.Adding strong acid or base normally means you will have H+ or OH- as a reactant, the rest is a spectator ion. All other acids/bases can be formed by reacting the conjugate ion with a strong acid/base.
8. Weak + Strong reactionReacting an acid with a base will produce water.HCl + NaOH H2O + NaClReacting a weak acid with a strong base will produce water and conjugate base.Sodium hydroxide and carbonic acidH2CO3 + OH- H2O + HCO3-Reacting a weak base with a strong acid will produce conjugate acid.Nitric acid and ammoniaNH3 + H+ NH4+
9. ExamplesCalcium hydroxide reacts with chlorous acidAcetic acid reacts with potassium hydroxideHydrochloric acid reacts with calcium nitriteNitric acid reacts with sodium chloriteHydroiodic acid reacts with sodium hydroxide
10. Acids and BasesWater is the product of all neutralization reactions between an acid and a baseH2O(l) ⇌ H+ (aq) + OH-(aq)The Arrhenius Definitions.An acid contains H and forms H+ (proton)When that H+ is donated to H2O it forms H3O+ (hydronium).A base contains OH and forms OH- (hydroxide).10
11. Brønsted-LoweryThe Brønsted-Lowery definitions, the proton transfer: An acid is a proton (H+) donor.A base is a proton (H+) acceptor.which is an acid/base?HF + H2O ⇌ H3O+ + F-NH3 + H2O ⇌ NH4+ + OH-11
12. Follow the protonHF + H2O ⇌ H3O+ + F-NH3 + H2O ⇌ NH4+ + OH-What about the reverse reaction?H+H+H+H+
13. Conjugate acids and basesWhen you run the reverse reaction you find the products are also acids and bases. The acids and bases that are formed are called conjugate acids or basesH2O + HF ⇌ H3O+ + F-base acid conjugate acid conjugate baseNH3 + H2O ⇌ NH4+ + OH-base acid CA CB
14. Label Acid, Base, Conjugate Acid, Conjugate BaseHClO3 + H2O ⇌ ClO3- +H3O+ A B CB CA ClO- + H2O ⇌ HClO + OH- B A CA CBHSO4- + H2O ⇌ SO42- +H3O+ A B CB CAAgOH + H2O ⇌ Ag+ + H2O + OH- B A CA CB
15. Ammonia SmellNH3 + H2O ⇌ NH4++OH-Ammonia is a gas with a distinct odorAmmonium and hydroxide are both odorless.What will happen if you add an acid or a base to this equilibrium?If base is added to the solution the equilibrium will shift to the left so you will smell ammonia, if hydroxide is removed (acid is added), it will shift to the right to so you won’t smell anything.
16. Pet “Stain” ProblemUrine has ammonia in it.Most cleansers are basic.After cleaning, we still leave small amounts behind.If it is small amount of ammonia and a basic cleanser the equilibrium will be shifted to the ammonia side so some thing with a great sense of smell (dog) could pick it up.A slightly acidic cleanser shifts the equilibrium to the ammonium side to solve this problem
17. Conjugate acids and bases …Conjugate acids and bases determine if an acid or base is strong or weak.If the conjugate acid/base readily reacts to run the reverse reaction it is a weak acid/base.If it does not react in the reverse reaction the acid or base is strong.
18. Acid Equilibrium constantFor some acid “A”HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq) Ka = [H3O+] [ A-] [HA]It is also writtenH A ⇌ H+ + A- Ka = [H+] [ A-] [HA]
19. ProblemsAcid Dissociation (Ionization ) Reactions.Write the simple dissociation (ionization) reaction for each of the following acids. Then write the Ka expressionHydrochloric acidAcetic acidThe ammonium ionThe annilium ion (C6H5NH3+)The hydrated aluminum(III) ion [Al(H2O)6]3+
20. Acid StrengthStrong acids dissociate completely in water.At equilibrium, Q = Ka >> 1 because [HA] is approx. 0.Weak acids are mostly undissociated.At equilibrium, Q = Ka << 1 because [H3O+] and [ A-] are very small.The smaller the Ka value, the weaker the acid.
21. Ka values of some weak acids21
22. Strong acidsStrong acids are not an equilibrium reaction.HCl + H2O H3O+ + Cl-Ka cannot be accurately determined in water because the reaction lies so far to the right that [HCl] is too small to measure.That is true for all strong acids, so Ka is normally only used for weak acids.
23. WaterWater is a stronger base than the conjugate acid of a strong base.Water would be weaker than the conjugate acid of a weak baseRemember water can act as an acid or a base.Amphoteric substances, like water, act as either an acid or a base
24. The Autoionization of Water and the Ion-Product (Dissociation)Kw is the constant for Water H2O(l) + H2O(l) ⇌ H3O+ (aq) + OH-(aq)At 25o C Kc = Kw = [H3O+] [OH-] = 1.0 x 10-14 if [H3O+] = [OH-] = 1.0 x 10-7M if [H3O+] > [OH-] = Acidic. if [H3O+] < [OH-] = Basic. if [H3O+] = [OH-] = Neutral.
25. Calculating [H3O+] and [OH-]Calculate [H3O+] or [OH-] as required for each of the following solutions at 25o C, and state whether the solution is neutral, acidic, or basic.1.0 x10-5 M OH-2.3 x10-7 M OH-10.0 M H3O+
26. ProblemAt 60o C, the value of Kw is 1.0 x10-13.Using LeChatelier’s principle, predict whether the reactionH2O(l) + H2O(l) ⇌ H3O+ (aq) + OH-(aq)is exothermic or endothermic. Calculate [H3O+] and [OH-] in a neutral solution at 60o C.
27. pH[H3O+] = 10-pH Equation sheet
28. Sig Figs and pHThe number of decimal places in the log value, pH value, is equal to the number of significant figures in the number that we took the logarithm of, concentration.Calculate pH and pOH for each of the following solutions.2.7 x 10-3 M OH-3.4 x 10-5 M H3O+1.54 x 10-10 M OH-
29. pH problemThe pH of a sample of human blood was measured to be 7.41 at 25o C. Calculate pOH, [H3O+], and [OH-] for the sample.