Zum notes (09) acids and bases

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  • Zum notes (09) acids and bases

    1. 1. by Steven S. Zumdahl & Donald J. DeCoste University of Illinois Introductory Chemistry: A Foundation, 6 th Ed. Introductory Chemistry, 6 th Ed. Basic Chemistry, 6 th Ed.
    2. 2. Chapter 16 Acids and Bases
    3. 3. Properties of Acids <ul><li>Sour taste </li></ul><ul><li>Change color of vegetable dyes (“indicators”) </li></ul><ul><li>React with “active” metals </li></ul><ul><ul><li>Like Al, Zn, Fe, but not Cu, Ag or Au </li></ul></ul><ul><ul><li>Zn + 2 HCl  ZnCl 2 + H 2 </li></ul></ul><ul><li>Corrosive </li></ul><ul><li>React with carbonates, producing CO 2 </li></ul><ul><ul><li>Marble, baking soda, chalk </li></ul></ul><ul><ul><li>CaCO 3 + 2 HCl  CaCl 2 + CO 2 + H 2 O </li></ul></ul><ul><li>React with bases to form ionic salts, and often water </li></ul>
    4. 4. Properties of Bases <ul><li>Also known as alkalis </li></ul><ul><li>Bitter Taste </li></ul><ul><li>Feel slippery </li></ul><ul><li>Change color of vegetable dyes </li></ul><ul><ul><li>Different color than acid </li></ul></ul><ul><ul><li>Litmus = blue </li></ul></ul><ul><li>React with acids to form ionic salts, and often water (HCl + NaOH  NaCl + HOH) </li></ul><ul><ul><li>Neutralization </li></ul></ul>
    5. 5. Bubbles (CO 2 ) NR Limestone CaCO 3 Bubbles (H 2 ) NR Magnesium Yellow Blue Bromothymol Stayed clear (cloudy) Clear  Pink (Fuschia) Phenolphthalein Blue  Red Red  Blue Litmus (blue or red) <7 >7 pH (# from the key) Not slippery Slippery Feel (choose slippery or not slippery) Sour Bitter Taste Acids ( H + + A - ) Bases (X + + OH - ) Lab Results
    6. 6. Mnemonic Device <ul><li>H + A - B + OH - </li></ul><ul><li>clea R O Y G B I V uschia </li></ul><ul><li>1 7 14 </li></ul><ul><li>BTB </li></ul><ul><li>universal </li></ul><ul><li>phenolphthalein </li></ul>pH = litmus
    7. 7. Arrhenius Theory <ul><li>Acids ionize in water to H + ions and anions </li></ul><ul><li>Bases ionize in water to OH - ions and cations </li></ul><ul><li>Neutralization reaction involves H + combining with OH - to make water </li></ul><ul><li>H + ions are protons </li></ul>
    8. 8. <ul><li>Definition only good in water solution </li></ul><ul><li>Definition does not explain why ammonia solutions turn litmus blue </li></ul><ul><ul><li>Basic without OH - ions </li></ul></ul>Arrhenius Theory (cont.)
    9. 9. Brønsted-Lowery Theory <ul><li>H + transfer reaction </li></ul><ul><ul><li>Since H + is a proton, also known as proton transfer reactions </li></ul></ul><ul><li>In the reaction, a proton from the acid molecule is transferred to the base molecule </li></ul><ul><li>Products are called the conjugate acid and conjugate base </li></ul>
    10. 10. Brønsted-Lowery Theory (cont.) <ul><li>H-A + :B  A - + H-B + </li></ul><ul><li>A - is the conjugate base, H-B + is the conjugate acid </li></ul><ul><li>Conjugate acid-base pair is either the original acid and its conjugate base or the original base and its conjugate acid </li></ul><ul><ul><li>H-A and A - are a conjugate acid-base pair </li></ul></ul><ul><ul><li>:B and H-B + are a conjugate acid-base pair </li></ul></ul>
    11. 11. Complete the following table
    12. 12. Complete the following table Produces H+ ions in solution Produces OH- ions in solution Proton donor Proton acceptor
    13. 13. Example #1: <ul><li>Determine what species you will get if you remove 1 H +1 from the acid. </li></ul><ul><ul><li>Conjugate base will have one more negative charge than the original acid </li></ul></ul><ul><ul><li>H 3 PO 4  H + + H 2 PO 4 - </li></ul></ul>Write the conjugate base for the acid H 3 PO 4
    14. 14. Brønsted-Lowery Theory (cont.) <ul><li>In this theory, instead of the acid, HA, dissociating into H + (aq) and A - (aq), the acid donates its H to a water molecule </li></ul><ul><li>HA + H 2 O  A - + H 3 O + </li></ul><ul><li>A -1 is the conjugate base </li></ul><ul><li>H 3 O + is the conjugate acid </li></ul>
    15. 15. Brønsted-Lowery Theory (cont.) <ul><li>H 3 O + is called the hydronium ion </li></ul><ul><li>In this theory, substances that do not have OH - ions can act as a base if they can accept a H +1 from water. </li></ul><ul><li>H 2 O + :B  OH - + H-B + </li></ul>
    16. 16. Acids in Water
    17. 17. Identify the acid and base in the following reaction acid base Conjugate acid (hydronium ion) Conjugate base Conjugate acid-base pair Conjugate acid-base pair
    18. 18. Bases in Water acid base Conjugate acid (hydronium ion) Conjugate base Conjugate acid-base pair Conjugate acid-base pair
    19. 19. Identify the conjugate acid-base pairs <ul><li>HSO 4 - + H 2 O  SO 4 2- + H 3 O + </li></ul><ul><li>HCO 3 - + H 2 O  OH - + H 2 CO 3 </li></ul><ul><li>HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2 - + H 3 O + </li></ul><ul><li>HPO 4 2- + H 2 O  H 2 PO 4 - + OH - </li></ul>
    20. 20. Identify the conjugate acid-base pairs <ul><li>HSO 4 - + H 2 O  SO 4 2 - + H 3 O + A B CB CA </li></ul><ul><li>HCO 3 - + H 2 O  OH - + H 2 CO 3 B CA CB CA </li></ul><ul><li>HC 2 H 3 O 2 + H 2 O  C 2 H 3 O 2 - + H 3 O + A B CB CA </li></ul><ul><li>HPO 4 2- + H 2 O  H 2 PO 4 - + OH - B A CA CB </li></ul>
    21. 21. Complete the following tables
    22. 22. Complete the following tables NO 3 - H 2 O OH - H 2 SO 4 Br - OH - H 3 O + HCO 3 - HSO 4 - ClO 4 -
    23. 23. Strength of Acids & Bases <ul><li>The stronger the acid, the more willing it is to donate H + (i.e. 100% dissociation) </li></ul>
    24. 24. Strength of Acids & Bases (cont.) <ul><li>Strong bases will react completely with water to form hydroxide: </li></ul><ul><li>CO 3 -2 + H 2 O  HCO 3 - + OH - </li></ul><ul><li>Only small fraction of weak base molecules pull H + off water: </li></ul><ul><li>HCO 3 - + H 2 O  H 2 CO 3 + OH - </li></ul>
    25. 25. Multiprotic Acids <ul><li>Monoprotic acids have 1 acid H, diprotic 2, etc. </li></ul><ul><ul><li>In oxyacids only the H on the O is acidic (why?) </li></ul></ul><ul><li>In strong multiprotic acids, like H 2 SO 4 , only the first H is strong; transferring the second H is usually weak </li></ul><ul><li>H 2 SO 4 + H 2 O  H 3 O + + HSO 4 - </li></ul><ul><li>HSO 4 - + H 2 O  H 3 O + + SO 4 -2 </li></ul>
    26. 26. What new words describe H 2 SO 4 ?
    27. 27. What new words describe H 2 SO 4 ? <ul><li>Strong acid  100% dissociated (i.e. strong electrolyte) </li></ul><ul><li>Diprotic (two protons that it can donate) </li></ul><ul><li>Oxyacid (contains oxygen) </li></ul><ul><li>Note: since it is a strong acid HSO 4 - is a weak base (strong acids always form weak conjugate bases) </li></ul>
    28. 28. Water As an Acid and a Base <ul><li>Amphoteric substances can act as either an acid or a base. </li></ul><ul><ul><li>Water as an acid, NH 3 + H 2 O  NH 4 + + OH - </li></ul></ul><ul><ul><li>Water as a base, HCl + H 2 O  H 3 O + + Cl - </li></ul></ul><ul><li>Water can even react with itself: </li></ul><ul><li>H 2 O + H 2 O  H 3 O + + OH - </li></ul>
    29. 29. Autoionization of Water <ul><li>Water is an extremely weak electrolyte. </li></ul><ul><ul><li>Therefore there must be a few ions present </li></ul></ul><ul><li>H 2 O + H 2 O  H 3 O + + OH - </li></ul>
    30. 30. * Acidic and Basic Solutions * <ul><li>Acidic solutions have a larger [H + ] than [OH - ] </li></ul><ul><li>Basic solutions have a larger [OH - ] than [H + ] </li></ul><ul><li>Neutral solutions have [H + ]=[OH - ]= 1 x 10 -7 M </li></ul><ul><li>K w = [H + ][OH - ]= 1 x 10 -14 </li></ul>[H + ] = 1 x 10 -14 [OH - ] [OH - ] = 1 x 10 -14 [H + ]
    31. 31. pH scale
    32. 32. Example #2 Determine the [H + ] and [OH - ] in a 10.0 M H + solution
    33. 33. Example #2 (cont.) <ul><li>Determine the given information and the information you need to find </li></ul><ul><ul><li>Given [H + ] = 10.0 M, find [OH - ] </li></ul></ul>
    34. 34. <ul><ul><li>Given [H + ] = 10.0 M = 1.00 x 10 1 M </li></ul></ul><ul><ul><li>K w = 1.0 x 10 -14 </li></ul></ul>Example #2 (cont.)
    35. 35. pH & pOH <ul><li>The acidity/basicity of a solution is often expressed as pH or pOH. </li></ul><ul><li>pH = -log[H 3 O + ] pOH = -log[OH - ] </li></ul><ul><ul><li>pH water = -log[10 -7 ] = 7 = pOH water </li></ul></ul><ul><li>[H + ] = 10 -pH [OH - ] = 10 -pOH </li></ul>
    36. 36. pH & pOH (cont.) <ul><li>pH < 7 is acidic; pH > 7 is basic, pH = 7 is neutral </li></ul><ul><li>The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution </li></ul><ul><li>1 pH unit corresponds to a factor of 10 difference in acidity </li></ul><ul><li>14 = pH + pOH </li></ul>
    37. 37. Example #3 Calculate the pH of a solution with a [OH - ] = 1.0 x 10 -6 M
    38. 38. Example #3 (cont.) <ul><li>Find the concentration of [H + ] </li></ul>
    39. 39. <ul><li>Enter the [H + ] concentration into your calculator and press the log key </li></ul><ul><ul><li>log(1.0 x 10 -8 ) = -8.0 </li></ul></ul><ul><li>Change the sign to get the pH </li></ul><ul><ul><li>pH = -(-8.0) = 8.0 </li></ul></ul>Example #3 (cont.)
    40. 40. <ul><li>Enter the [H + ] or [OH - ] concentration into your calculator and press the log key </li></ul><ul><ul><li>log(1.0 x 10 -3 ) = -3.0 </li></ul></ul><ul><li>Change the sign to get the pOH </li></ul><ul><ul><li>pOH = -(-3) = 3.0 </li></ul></ul><ul><li>Subtract the calculated pH or pOH from 14.00 to get the other value </li></ul><ul><ul><li>pH = 14.00 – 3.0 = 11.0 </li></ul></ul>Example #4 Calculate the pH and pOH of a solution with a [OH - ] = 1.0 x 10 -3 M
    41. 41. <ul><li>If you want to calculate [OH - ] use pOH; if you want [H + ] use pH. It may be necessary to convert one to the other using 14 = [H + ] + [OH - ] </li></ul><ul><ul><li>pOH = 14.00 – 7.41 = 6.59 </li></ul></ul>Example #5 Calculate the [OH - ] of a solution with a pH of 7.41
    42. 42. Example #5 (cont.) <ul><li>Enter the pH or pOH concentration into your calculator </li></ul><ul><li>Change the sign of the pH or pOH </li></ul><ul><ul><li>-pOH = -(6.59) </li></ul></ul><ul><li>Press the button(s) on you calculator to take the inverse log or 10 x </li></ul><ul><ul><li>[OH - ] = 10 -6.59 = 2.6 x 10 -7 M </li></ul></ul>
    43. 43. Calculating the pH of a Strong, Monoprotic Acid <ul><li>A strong acid will dissociate 100% </li></ul><ul><li>HA  H + + A - </li></ul><ul><li>Therefore the molarity of H + ions will be the same as the molarity of the acid </li></ul><ul><li>Once the H + molarity is determined, the pH can be determined </li></ul><ul><li>pH = -log[H + ] </li></ul>
    44. 44. Example #6 Calculate the pH of a 0.10 M HNO 3 solution. pH means – log of [H+] pH = - log [0.10] pH = - log [ 1 x 10 -1 ] pH = 1 note the exponent!
    45. 45. Example #6 (cont.) <ul><li>Determine the [H + ] from the acid concentration </li></ul><ul><ul><li>HNO 3  H + + NO 3 - </li></ul></ul><ul><ul><li>0.10 M HNO 3 = 0.10 M H + </li></ul></ul><ul><li>Enter the [H + ] concentration into your calculator and press the log key </li></ul><ul><ul><li>log(0.10) = -1.00 </li></ul></ul><ul><li>Change the sign to get the pH </li></ul><ul><ul><li>pH = -(-1.00) = 1.00 </li></ul></ul>
    46. 46. Practice Problems <ul><li>A HCl solution is 8.34 x 10 -5 mole/liter. Estimate, then calculate the pH of the solution. </li></ul><ul><li>What is the [OH - ] of a solution whose pOH = 2.86 ? </li></ul><ul><li>What is the [OH - ] of a solution whose [H + ] = 0.001M </li></ul><ul><li>The pH of a soft drink is determined to be 4.0. What is the [OH - ] of the drink? </li></ul><ul><li>What is the pH of a 0.001 M Mg(OH) 2 solution? (Assume 100% dissociation) </li></ul>
    47. 47. Practice Problems <ul><li>A HCl solution is 8.34 x 10 -5 mole/liter. Estimate, then calculate the pH of the solution. </li></ul><ul><li>pH < 5 (see the exponent) </li></ul><ul><li>pH = - log [H+] </li></ul><ul><li>pH = - log[8.34 x 10 -5 ] </li></ul><ul><li>pH = 4.079 (pH < 7, acidic) </li></ul>
    48. 48. Practice Problems <ul><li>What is the [OH - ] of a solution whose pOH = 2.86 ? </li></ul><ul><li>pOH means – log[OH-] </li></ul><ul><li>-log[OH-] = 2.86 </li></ul><ul><li>log[OH-] = - 2.86 </li></ul><ul><li>[OH-] = 10 -2.86 </li></ul><ul><li>[OH-] = 0.0014 M since [OH-]>[H+] solution is basic </li></ul>
    49. 49. Practice Problems <ul><li>What is the [OH - ] of a solution whose [H + ] = 0.001M </li></ul><ul><li>[H+][OH-] = 1 x 10 -14 </li></ul><ul><li>[0.001][OH-] = 1 x 10 -14 </li></ul><ul><li>[OH-] = 1 x 10 -11 </li></ul><ul><li>Since [OH-] < [H+] solution is acidic </li></ul>
    50. 50. Practice Problems <ul><li>The pH of a soft drink is determined to be 4.0. What is the [OH - ] of the drink? </li></ul><ul><li>pH + pOH = 14 </li></ul><ul><li>4 + pOH = 14 </li></ul><ul><li>pOH = 10 </li></ul><ul><li>-log [OH] = 10 </li></ul><ul><li>[OH-] = 10 -10 since [OH-] < [H+] soda is acidic </li></ul>
    51. 51. Practice Problems <ul><li>What is the pH of a 0.001 M Mg(OH) 2 solution? (Assume 100% dissociation) </li></ul><ul><li>[OH-] = 2 * 0.001 </li></ul><ul><li>pOH = - log [0.002] </li></ul><ul><li>pOH = 2.699 </li></ul><ul><li>pH + pOH = 14 </li></ul><ul><li>pH = 14 – 2.699 = 11.30 (pH > 7, basic) </li></ul>
    52. 52. Challenge Questions <ul><li>What is the pOH of a 0.0025 M acetic acid solution that is only 8.5% dissociated? </li></ul><ul><li>How much more acidic is a solution whose pH is 6.0 compared to a solution whose pH is 12.0? </li></ul><ul><li>What is the resulting pH if equal volumes of solutions are mixed, one with a pH of 6.0 and one with a pH of 12.0? </li></ul>
    53. 53. Challenge Questions <ul><li>What is the pOH of a 0.0025 M acetic acid solution that is only 8.5% dissociated? </li></ul><ul><li>HA  H+ + A- 0.0025 0 0 - 2.125 x 10 -4 + 2.125 x 10 -4 +2.125 x 10 -4 </li></ul><ul><li>0.0025 * 0.085 = 2.125 x 10 -4 M [H+] </li></ul><ul><li>14 – (-log(2.125 x 10 -4 ) = </li></ul>
    54. 54. Challenge Questions <ul><li>How much more acidic is a solution whose pH is 6.0 compared to a solution whose pH is 12.0? </li></ul><ul><li>pH = 6, [H+] = 0.000001 </li></ul><ul><li>pH = 12 [H+] = 0.000000000001 </li></ul>
    55. 55. Challenge Questions <ul><li>What is the resulting pH if equal volumes of solutions are mixed, one with a pH of 6.0 and one with a pH of 12.0? </li></ul><ul><li>pH = 6, [H+] = 0.000001 </li></ul><ul><li>pH = 12 [H+] = 0.000000000001 </li></ul><ul><li>(0.000001 + 0.000000000001)/2 = </li></ul><ul><li>- log ( ) = </li></ul>
    56. 56. Titration <ul><li>Laboratory Set-up </li></ul><ul><li>Sample Problem </li></ul><ul><li>Determine the unknown concentration of HCl if 25.0 mL of the acid are neutralized with 50.0 mL of 0.100 M NaOH. </li></ul><ul><li>H + + OH -  HOH </li></ul>
    57. 57. Practice Problems <ul><li>Calculate the volume of 0.300 M HCl needed to titrate 75.00 mL of 0.1500 M KOH (aq) . </li></ul><ul><li>Determine the volume of 0.100 M NaOH needed to reach the equivalence (end) point against 50.0 mL of 0.200 M HNO 3. </li></ul>
    58. 58. Buffered Solutions <ul><li>Buffered solutions resist change in pH when an acid or base is added to it. </li></ul><ul><li>Used when need to maintain a certain pH in the system </li></ul><ul><ul><li>Blood </li></ul></ul>
    59. 59. Buffered Solutions (cont.) <ul><li>A buffer solution contains a weak acid and its conjugate base. </li></ul><ul><li>Buffers work by reacting with added H + or OH - ions so they do not accumulate and change the pH. </li></ul><ul><li>Buffers will only work as long as there are sufficient weak acid and conjugate base molecules present. </li></ul>
    60. 60. Challenge Question <ul><li>Give two components of a buffer. </li></ul><ul><li>Identify which component will react with added acid, show using a balanced equation. </li></ul><ul><li>Identify which component will react with added base, show using a balanced equation. </li></ul>

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