1 / 31

Acids and Bases

Acids and Bases. Acids taste sour (citric acid, acetic acid) Bases taste bitter (sodium bicarbonate) There are 3 ways to define acids and bases, you will learn 2 of these: Arrhenius: - Acids form H 3 O + in water (HCl + H 2 O  H 3 O + + Cl - )

shelby-fuentes
Télécharger la présentation

Acids and 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. Acids and Bases • Acids taste sour (citric acid, acetic acid) • Bases taste bitter (sodium bicarbonate) • There are 3 ways to define acids and bases, you will learn 2 of these: Arrhenius: - Acids form H3O+ in water (HCl + H2O  H3O+ + Cl-) - Bases form OH- in water (NaOH  Na+ + OH-) Brønsted-Lowry (B-L): - Acids donate H+ and Bases accept H+ HCl + NaOH  H2O + NaCl HCl is the acid, it donates H+ to OH- (the base)

  2. B-L Acids and Formation of H3O+ • In an acid-base reaction, there is always an acid/base pair (the acid donates H+ to the base) • H+ is not stable alone, so it will be transferred from one covalent bond to another Example: formation of H3O+ from an acid in water HBr + H2O  H3O+ + Br-

  3. Identifying B-L Acids and Bases • Compare the reactants and the products - The reactant that loses an H+ is the acid - The reactant that gains an H+ is the base Examples: HCl + H2O  H3O+ + Cl- Acid = HCl and Base = H2O (HCl gives H2O an H+) NH3 + H2O  NH4+ + OH- Acid = H2O and Base = NH3 (H2O gives NH3 an H+) CH3CO2H + NH3 CH3CO2- + NH4+ Acid = CH3CO2H and Base = NH3 (CH3CO2H gives NH3 an H+)

  4. Conjugate Acids and Bases • When a proton is transferred from the acid to the base (in a B-L acid/base reaction), a new acid and a new base are formed: HA + B  A- + HB+ acid + base  conjugate base + conjugate acid • The acid (HA) and the conjugate base (A-) that forms when HA gives up an H+ are a conjugate acid/base pair • The base (B) and the conjugate acid (HB+) that forms when B accepts an H+ are another conjugate acid/base pair

  5. Identifying Conjugate Acid/Base Pairs • Identify the acid and base for the reactants • Identify the acid and base for the products • Identify the conjugate acid/base pairs acid conjugate base + + baseconjugate acid HF H3O+ F- H2O

  6. Acid and Base Strength • Strong acids give up protons easily and completely ionize in water: HCl + H2O  H3O+ + Cl- • Weak acids give up protons less easily and only partially ionize in water: CH3CO2H + H2O  CH3CO2- + H3O+ • Strong bases have a strong attraction for H+ and completely ionize in water: KOH(s)  K+ (aq) + OH-(aq) NaNH2 + H2O  NH3 + NaOH • Weak bases have a weak attraction for H+ and only partially ionize in water: HS- + H2O  H2S + OH-

  7. Direction of an Acid/Base Equilibrium • In general, there’s an inverse relationship between acid/base strength within a conjugate pair: - strong acid  weak conjugate base - strong base  weak conjugate acid (and vice-versa) • The equilibrium always favors the direction that goes from stronger acid to weaker acid Example 1: HBr + H2O ? H3O+ + Br- stronger acid (HBr)  weaker acid (H3O+) (equilibrium favors products) Example 2: NH3 + H2O ? NH4+ + OH- weaker acid (H2O)  stronger acid (NH4+) (equilibrium favors reactants)

  8. Dissociation Constants • Since weak acids dissociate reversibly in water, we can write an equilibrium expression: HA + H2O  H3O+ + A- Keq = [H3O+][A-]/[HA][H2O] • But, since [H2O] remains essentially constant we can write: Ka = Keq x [H2O] = [H3O+][A-]/[HA] • The acid dissociation constant (Ka) is a measure of how much the acid dissociates (A higher Ka = a stronger acid) • Example: CH3CO2H + H2O  CH3CO2- + H3O+ Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5 • Can also write dissociation constants for weak bases: NH3 + H2O  NH4+ + OH- Kb = [NH4+][OH-]/[NH3] = 1.8 x 10-5

  9. Ionization of Water • Since H2O can act as either a weak acid or a weak base, one H2O can transfer a proton to another H2O: H2O + H2O  H3O+ + OH- Keq = [H3O+][OH-]/[H2O][H2O] • Since [H2O] is essentially constant, we can write: Kw = Keq x [H2O]2 = [H3O+][OH-] (where Kw = the ion-product constant for water) • For pure water: [H3O+] = [OH-] = 1.0 x 10-7 M So, Kw = [H3O+][OH-] = (1.0 x 10-7 M)2 = 1.0 x 10-14 (units are omitted for Kw as for Keq and Ka)

  10. Using Kw • If acid is added to water, [H3O+] goes up - for an acidic solution [H3O+] > [OH-] • If base is added to water, [OH-] goes up - for a basic solution [OH-] > [H3O+] • Kw is constant (1.0 x 10-14) for all aqueous solutions • Can use Kw to calculate either [H3O+] or [OH-] if given the other concentration • Example: if [H3O+] = 1.0 x 10-4 M, what is the [OH-]? Kw = [H3O+][OH-] [OH-] = Kw/ [H3O+] = 1.0 x 10-14/1.0 x 10-4= 1.0 x 10-10 M Is this an acidic or a basic solution? Since [H3O+] > [OH-], it’s an acidic solution

  11. The pH Scale • pH is a way to express [H3O+] in numbers that are easy to work with • [H3O+] has a large range (1.0 M to 1.0 x 10-14 M) so we use a log scale: pH = - log [H3O+] • The pH scale goes from 0 - 14 • Each pH unit = a ten-fold change in [H3O+] • pH 7 = neutral, pH < 7 = acidic, pH > 7 = basic • Can use an indicator dye (on paper or in solution) that changes color with changes in pH, or a pH meter, to measure pH

  12. Calculating pH and pOH • Can calculate pH from [H3O+]: If [H3O+] = 1.0 x 10-3 M, what is the pH? pH = - log [H3O+] = - log(1.0 x 10-3) = 3.00 • Note: sig. figs. in [H3O+] = decimal places in pH • Can also calculate [H3O+] from pH: If pH is a whole number, [H3O+] = 1 x 10-pH So, if pH = 2, then [H3O+] = 1 x 10-2 • Can calculate pOH from [OH-] pOH = - log[OH-] • Also, since Kw = [H3O+][OH-] then pKw = - log Kw = - log (1.0 x 10-14) = 14.00 • And, pKw = pH + pOH = 14.00 • So, if pH = 3.00, then pOH = 14.00 - 3.00 = 11.00

  13. Reactions of Acids and Bases • Acids and bases are involved in a variety of chemical reactions (we’ll study 3 types here) • Acids react with certain metals to produce metal salts and H2 gas, for example: Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) • Acids react with carbonates and bicarbonates to produce salts, H2O and CO2 gas, for example: NaHCO3(aq) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g) • Acids react with bases (neutralization reactions) to form salts and H2O, for example: HBr(aq) + LiOH(aq)  LiBr (aq) + H2O(l) • Neutralization reactions are balanced with respect to moles of H+ and moles of OH-, for example: H2SO4 + 2NaOH  Na2SO4 + 2H2O

  14. Acidity of Salt Solutions • Salts dissolved in water can affect the pH • When salts dissolve, they dissociate into their ions NaCl  Na+ + Cl- • If one of those ions can donate a proton to H2O, or accept one from H2O, the pH will change: Na2S  2Na+ + S2- S2- + H2O  HS- + OH- S2- is a weak base that can accept an H+ from H2O Since [OH-] is increased, the solution is basic

  15. Salts that form Neutral Solutions • When a strong acid dissolves in water, a weak conjugate base is formed that can’t remove a proton from water • When a strong base dissolves in water, the metal that dissociates can’t form H3O+ • So, salts containing ions that come from strong acids and bases do not affect the pH of the solution • Example: KBr  K+ + Br- (KOH = strong base, HBr = strong acid) (KOH + HBr  KBr + H2O) K+ has no proton to donate, so can’t form H3O+ Br- is too weak of a base to pull a proton off of H2O, so can’t form OH- So, the solution remains neutral

  16. Salts that form Basic Solutions • When a weak acid dissolves in water, the conjugate base formed is usually strong enough to remove a proton from H2O to form OH- • So, salts that contain ions that come from a weak acid and a strong base form basic solutions • Example: NaCN  Na+ + CN- (HCN is a weak acid) CN- + H2O  HCN + OH- (HCN + NaOH  NaCN + H2O) (Na+ doesn’t affect the pH, it’s from a strong base)

  17. Salts that form Acidic Solutions • When a weak base dissolves in water, the conjugate acid formed is usually strong enough to donate a proton to H2O to form H3O+ • So, salts that contain ions from a weak base and a strong acid form acidic solutions • Example: NH4Br  NH4+ + Br- (from NH3 and HBr) (NH3 + HBr  NH4Br) NH4+ + H2O  NH3 + H3O+ (Br- doesn’t affect the pH)

  18. Buffer solutions • A small amount of strong acid or base added to pure water will cause a very large change in pH • A buffer is a solution that can resist changes in pH upon addition of small amounts of strong acid or base • Body fluids, such as blood, are buffered to maintain a fairly constant pH • Buffers are made from conjugate acid/base pairs (either a weak acid and a salt of its conjugate base or a weak base and a salt of its conjugate acid) • Thus, they contain an acid to neutralize any added base, and a base to neutralize any added acid • Buffers can’t be made from strong acids or bases and the salts of their conjugates since they completely ionize in H2O

  19. How to Make a Buffer Solution • An acetate buffer is made from acetic acid and a salt of its conjugate base: CH3CO2H and CH3CO2Na • The salt is used to increase the concentration of CH3CO2- in the buffer solution • Recall: CH3CO2H + H2O  CH3CO2- + H3O+ (the equilibrium favors reactants, so the concentration of CH3CO2- is low) But, CH3CO2Na  CH3CO2- + Na+ CH3CO2H + CH3CO2Na + H2O  2 CH3CO2- + H3O+ + Na+ • If acid is added: CH3CO2- + H3O+  CH3CO2H + H2O • If base is added: CH3CO2H + OH-  CH3CO2- + H2O • Buffer capacity = how much acid or base can be added and still maintain pH (depends on buffer type and concentration)

  20. Calculating pH of a Buffer • The pH of a buffer solution can be calculated from the acid dissociation constant (Ka) • Example (for acetate buffer): CH3CO2H + H2O  CH3CO2- + H3O+ Ka = [H3O+][CH3CO2-]/[CH3CO2H] = 1.8 x 10-5 [H3O+] = Ka x [CH3CO2H]/[CH3CO2-] • What is the pH of an acetate buffer that is 1.0 M CH3CO2H and 0.50 M CH3CO2Na? [H3O+] = 1.8 x 10-5 x 1.0 M/0.50 M = 3.6 x 10-5 M pH = - log[H3O+] = - log(3.6 x 10-5) = 4.44

  21. Dilutions • Often solutions are obtained and stored as highly concentrated stock solutions that are diluted for use (i.e. cleaning products, frozen juices) • When a solution is diluted by adding solvent, the volume increases, but amount of solute stays the same, so the concentration decreases: Mol solute = concentration (mol/L) x V (L) = constant So, C1V1 = C2V2 • For molarity, it becomes: M1V1 = M2V2

  22. Dilution Calculations • Example 1: What volume (in mL) of 8.0 M HCl is needed to prepare 1.0 L of 0.50 M HCl? M1V1 = M2V2 V1 = M2V2/ M1 V1 = 0.50 M x 1.0 L/ 8.0 M = 0.0625 L = 63 mL • Example 2: How many L of water do you need to add to dilute 0.50 L of a 10.0 M NaOH solution to 1.0 M ? V2 = M1V1/ M2 V2 = 10.0 M x 0.50 L/ 1.0 M = 5.0 L volume of water needed = 5.0 L - 0.50 L = 4.5 L

  23. Acid-Base Titration • Molarity of an acid or base solution of unknown concentration can be determined by titration: • A measured volume of the unknown acid or base is placed in a flask and a few drops of indicator dye (such as phenolpthalein) are added • A buret is filled with a measured molarity of known base or acid (the “titrant”) and small amounts are added until the solution changes color (neutralization endpoint) • At neutralization endpoint [H3O+] = [OH-] • Molarity of unknown is calculated from moles of titrant added (mole ratio comes from balanced chemical equation)

  24. Example: Titration of H2SO4 with NaOH • What is the molarity of a 10.0 mL sample of H2SO4 if the neutralization endpoint is reached after adding 15.0 mL of 1.00 M NaOH? • Calculate moles NaOH added: • 15.0 mL x (1 L/ 1000 mL) x (1.00 mol/ 1 L) = 0.0150 mol NaOH • Write the balanced chemical equation: H2SO4 + 2NaOH  Na2SO4 + 2H2O • Calculate moles H2SO4 neutralized: 0.0150 mol NaOH x 1 mol H2SO4/ 2 mol NaOH = 0.00750 mol H2SO4 • Calculate molarity of H2SO4: 0.00750 mol H2SO4/ 0.0100 L = 0.750 M H2SO4

More Related