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Acid and Base Equilibrium

Acid and Base Equilibrium. Some Properties of Acids. Produce H 3 O + ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) Taste sour Corrode metals Electrolytes React with bases to form a salt and water pH is less than 7

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Acid and Base Equilibrium

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  1. Acid and BaseEquilibrium

  2. Some Properties of Acids • Produce H3O+ ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) • Taste sour • Corrode metals • Electrolytes • React with bases to form a salt and water • pH is less than 7 • Turns blue litmus paper to red

  3. Some Properties of Bases • Produce OH- ions in water • Taste bitter, chalky • Are electrolytes • Feel soapy, slippery • React with acids to form salts and water • pH greater than 7 • Turns red litmus paper to blue “Basic Blue”

  4. Acid/Base definitions Definition 1: Arrhenius Arrhenius acid is a substance that produces (H3O+) in water Arrhenius base is a substance that produces OH- in water

  5. Acid/Base Definitions • Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom that has lost it’s electron!

  6. A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor conjugatebase conjugateacid base acid

  7. ACID-BASE THEORIES The Brønsted definition means NH3 is aBASE in water — and water is itself anACID

  8. Conjugate Pairs

  9. Learning Check! Label the acid, base, conjugate acid, and conjugate base in each reaction: HCl + OH-Cl- + H2O Acid Base Conj.Base Conj.Acid H2O + H2SO4   HSO4- + H3O+ Conj.Base Conj.Acid Base Acid

  10. Acids & Base Definitions Definition #3 – Lewis Lewis acid - a substance that accepts an electron pair Lewis base - a substance that donates an electron pair

  11. Lewis Acids & Bases Formation ofhydronium ion is also an excellent example. • Electron pair of the new O-H bond originates on the Lewis base.

  12. Lewis Acid/Base Reaction

  13. More About Water Autoionization Kw = [H3O+] [OH-] = 1.00 x 10-14 at 25 oC In a neutral solution [H3O+] = [OH-] and so [H3O+] = [OH-] = 1.00 x 10-7 M

  14. More About Water H2O can function as both an ACID and a BASE. In pure water there can beAUTOIONIZATION Equilibrium constant for water = Kw Kw = 1.00 x 10-14 = [H3O+] [OH-] at 25 oC Take -logs of both sides 14 = pH + pOH

  15. The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H3O+(or OH-) ion.Under 7 = acid 7 = neutral Over 7 = base

  16. Calculating the pH pH = - log [H3O+] ((Remember that the [ ] mean Molarity or concentration) Example: If [H3O+] = 1 X 10-10 pH = - log 1 X 10-10 pH = - (- 10) pH = 10 Example: If [H3O+] = 1.8 X 10-5, what is the pH? pH = - log 1.8 X 10-5 pH = - (- 4.74) pH = 4.74

  17. pH calculations – Solving for H3O+ If the pH of Coke is 3.12, what is the [H3O+] ? Because pH = - log [H3O+] then - pH = log [H3O+] Take antilog (10x) of both sides and get 10-pH =[H3O+] [H3O+] = 10-3.12 = 7.58 x 10-4 M *** to find antilog on your calculator, look for “Shift” or “2nd function” and then the log button

  18. Strong and Weak Acids/Bases • Generally acids and bases are divided into STRONG or WEAK ones. • Strong acids – 6 (HNO3, HCl, HBr, HI, HClO4, H2SO4) all others are weak • Strong Bases – group 1 and 2 hydroxides (all others are weak) except Be(OH)2

  19. Strong acids and bases • Strong acids and bases - are 100 % ionized. • No equilibrium is set up • [Acid] = [H30+] ( 1:1 ratio) • ACID H A + H2O  H3O+ + A- • BASE  B + H2O  BH+ + OH- Find the pH of these: • A 0.15 M solution of Hydrochloric acid 2) A 3.00 X 10-7 M solution of Nitric acid 3) What is the pH of 0.0034M H2SO4? pH = - log [H3O+] pH = - log 0.15 pH = - (- 0.82) pH = 0.82 pH = - log 3 X 10-7 pH = - (- 6.52) pH = 6.52 Finding pH

  20. pH of strong bases What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10-3 M) pOH = - log 0.0010 pOH = 3 pH = 14 – 3 = 11 OR Kw = [H3O+] [OH-] [H3O+] = 1.0 x 10-11 M pH = - log (1.0 x 10-11) = 11.00

  21. What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution? HNO3(aq) + H2O (l) H3O+(aq) + NO3-(aq) Ba(OH)2(s) Ba2+(aq) + 2OH-(aq) What is the pH of a 2 x 10-3 M HNO3 solution? HNO3 is a strong acid – 100% dissociation. 0.0 M 0.0 M Start 0.002 M 0.0 M 0.002 M 0.002 M End pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7 Ba(OH)2 is a strong base – 100% dissociation. 0.0 M 0.0 M Start 0.018 M 0.0 M 0.018 M 0.036 M End pH = 14.00 – pOH = 14.00 - (-log(0.036) )= 12.56 15.4

  22. Weak Acids/Bases • Weak acidsare much less than 100% ionized in water. • Equilibrium is set up. Common weak acids are acetic acid and weak base is ammonia

  23. Equilibria Involving Weak Acids and Bases Consider acetic acid, HC2H3O2 (HOAc) HC2H3O2(aq)+ H2O(l) H3O+(aq)+ C2H3O2–(aq) Acid Conj. base (K is designated Ka for ACID) K gives the ratio of ions (split up) to molecules (don’t split up)

  24. Ionization Constants for Acids/Bases Conjugate Bases Acids Increase strength Increase strength

  25. Equilibrium Constants for Weak Acids Weak acid has Ka < 1 Leads to small [H3O+] and a pH of 2 - 7

  26. Equilibrium Constants for Weak Bases Kb – base dissociation constant and is temp dependent Weak base has Kb < 1 Leads to small [OH-] and a pH of 12 - 7

  27. Relation of Ka, Kb, [H3O+] and pH

  28. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 1.Define equilibrium concs. in ICE table. [HOAc] [H3O+] [OAc-] initial change equilib 1.00 0 0 -x +x +x 1.00-xx x

  29. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 2.Write Ka expression This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: 10-5 or smaller is ok)

  30. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 3.Solve Ka expression First assume x is very small because Ka is so small. Now we can more easily solve this approximate expression.

  31. Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H3O+, OAc-, and the pH. Step 3.Solve Kaapproximateexpression x =[H3O+] = [OAc-] = 4.2 x 10-3 M pH = - log [H3O+] = -log (4.2 x 10-3) =2.37

  32. Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO2H. HCO2H + H2O  HCO2- + H3O+ Ka = 1.8 x 10-4 Approximate solution [H3O+] = 4.2 x 10-4 M,pH = 3.37 Exact Solution [H3O+] = [HCO2-] = 3.4 x 10-4 M [HCO2H] = 0.0010 - 3.4 x 10-4 = 0.0007 M pH = 3.47

  33. Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 Step 1.Define equilibrium concs. in ICE table [NH3] [NH4+] [OH-] initial change equilib 0.010 0 0 -x +x +x 0.010 - x x x

  34. Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 Step 2.Solve the equilibrium expression Assume x is small, so x = [OH-] = [NH4+] = 4.2 x 10-4 M and [NH3] = 0.010 - 4.2 x 10-4 ≈ 0.010 M The approximation is valid !

  35. Equilibria Involving A Weak Base You have 0.010 M NH3. Calc. the pH. NH3 + H2O  NH4+ + OH- Kb = 1.8 x 10-5 Step 3.Calculate pH [OH-] = 4.2 x 10-4 M so pOH = - log [OH-] = 3.37 Because pH + pOH = 14, pH = 10.63

  36. Percent Ionization for Weak Acids • Most weak acids ionize < 50% • Percent ionization (p) • General Weak Acid:HA(aq) + H2O(l) H3O+ (aq) + A- (aq) • p varies depending on concentration: increase [HA ] decreases p • This is caused by Le Chatelier’s Principle • Remember, for strong acids we assume complete ionization (100%)

  37. Examples The pH of a 0.10mol/L methanoic acid (HCOOH) solution is 2.38. Calculate the percent ionization of methanoic acid Ans: 4.2% Calculate the acid ionization constant (Ka )of acetic acid if a 0.1000mol/L solution at equilibrium at SATP has a percent ionization of 1.3% (Hint: ICE table Ans: 1.7x10-5

  38. Relationship Between Ka and Kb for Conjugate Base Pairs • Recall: Conjugate Pairs – an acid and base that differ by one hydrogen • Lets consider the hypothetical weak acid, HA, and its conjugate base, A - • HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)

  39. Relationship Between Ka and Kb for Conjugate Base Pairs • Now consider the hypothetical weak base, A- in water • A- (aq) + H2O (l)  HA (aq) + OH- (aq) • Now let’s put that together • HA (aq) + H2O (l)  H3O+ (aq) + A- (aq)  Ka • A- (aq) + H2O (l)  HA (aq) + OH- (aq)  Kb • Add both equations • 2H2O (l)  H3O+ (aq) + OH- (aq) •  Kw

  40. Relationship Between Ka and Kb for Conjugate Base Pairs Recall: Autoionization of water H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) Kw=1.00x10-14**must remember this value**

  41. Relationship Between Ionization Constants for Conjugate Base Pairs • For acids and bases whose chemical formulas differ by only one hydrogen (conjugate pairs) the following apply: • Kw = Ka x Kb • Kb =Kw/Ka • Ka = Kw/Kb • Therefore if only the Ka value is available in the table, we can determine the conjugate pairs Kb by using the above equations • Note: these equations show the larger the Ka the smaller the Kb • Stronger acid  weaker conjugate base • Weaker acid  stronger conjugate base

  42. Learning check! 1. What is the value of the base ionization constant (Kb) for the acetate ion, C2H3O2- (aq) Ans: 5.6x10-10 2. Calculate the percent ionization of propanoic acid, HC3H5O2(aq), if a 0.050 mol/L solution has a pH of 2.78 Ans. 3.3%

  43. Rapid changes in pH can kill fish and other organisms in lakes and streams. Soil pH is affected and can kill plants and create sinkholes

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