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Acids and Bases

Acids and Bases. Properties of Acids and Bases. Pg 236. Why do Acids and bases change . Arrhenius Definition. Acid : produces H + (or H 3 O + ) when dissolved in water  HCl(aq)  H + (aq) + Cl - (aq) H + (aq) + H 2 O(l)  H 3 O + (aq) NOTE: H 3 O + = hydronium ion

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Acids and Bases

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  1. Acids and Bases

  2. Properties of Acids and Bases Pg 236

  3. Why do Acids and bases change

  4. Arrhenius Definition • Acid: produces H+ (or H3O+) when dissolved in water  HCl(aq)  H+(aq) + Cl-(aq) H+(aq) + H2O(l)  H3O+(aq) NOTE: H3O+ = hydronium ion • Base: produces OH- when dissolved in water. NaOH(s)  Na+(aq) + OH-(aq)

  5. Bronsted-Lowry Definition Acids: proton (H+) donors HF(aq)  H+(aq) + F-(aq) H+(aq) + H2O(l)  H3O+(aq) Bases: proton acceptors NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) H2O: acts as an acid and a base = AMPHOTERIC

  6. Strength of Acids Strong Acids:  ionize (splits up into ions) almost 100% in water  mostly ions in solution  amount of HCl present is negligeable HCl(aq)  H+(aq) + Cl-(aq) Weak acids:  ionize poorly in water  not many of these ions present in solution  mostly acetic acid (HC2H3O2) HC2H3O2(aq) C2H3O2-(aq) + H+(aq) NOTE: strong acids are strong electrolytes and will conduct electricity better than weak acids.

  7. Strength of Bases Strong Bases: ionize almost 100% in water NaOH(s)  Na+(aq) + OH-(aq) Weak Bases: ionize poorly in water NH3(l) + H2O(l) NH4+(aq) + OH-(aq) NOTE: strong bases are strong electrolytes

  8. Conjugate Acids and Conjugate Bases • these differ by only one proton • Examples HCl Cl- SO42- HSO4- Lose a proton Acid Conjugate base of HCl Gain a proton Base Conjugateacid of SO42-

  9. Reactions with Water Conjugate acid-base pair: CH3CO2H/CH3CO2- Conjugate acid-base pair: H2O/H3O+

  10. Monoprotic, Diprotic and Triprotic Monoprotic ·donates one acidic proton ·eg: HCl + H2O  H3O+ + Cl- ·only one H+ to donate Diprotic ·donates two acidic protons ·eg: H2SO4 + H2O  H3O+ + HSO4- · HSO4- + H2O  H3O+ + SO42- Triprotic ·donates three acidic protons ·eg: H3PO4 + H2O  H3O+ + H2PO4- ·three H+ to donate

  11. Homework • Pg 251 #1, 2 • Pg 253 #4, 5, 6

  12. pH < 7 acidic pH = 7 neutral pH > 7 basic

  13. pH • a measure of acid strength • By definition all acids contain at least one acidic proton = H+ • HA is a symbol used to represent any general acid HA H+(aq) + A-(aq) H+ + H2O  H3O+ •  [H+] = [H3O+] • If a lot of H3O+ is produced the solution is very acidic. • pH is directly related to [H3O+].

  14. Self-Ionization of Water H2O(l) + H2O(l)  H3O+(aq) + OH-(aq) This reaction does not occur to any great extent. [H3O+] = 1 x 10-7 mol/L [OH-] = 1 x 10-7 mol/L Because both concentrations are equal water is said to be neutral. Therefore, if [H3O+] = [OH-] neutral [H3O+] > [OH-] acidic [H3O+] < [OH-] basic NOTE: [H3O+][OH-] = 1.0 x 10-14

  15. pH = -log[H3O+] • Expressing hydronium concentrations in scientific notation isn’t very convenient. The pH scale was developed to make the expression of H3O+ concentration more convenient. • [H3O+] is the concentration in mol/L

  16. Example 1: pH of Water The concentration of H3O+ is 1.0 x 10-7. Calculate the pH.  pH = -log[H3O+] = -log(1.0 x 10-7) = -(-7) = 7 The pH of water is 7. Therefore pH 7 is neutral.

  17. Example 2 Determine the pH of a 1M solution of HCl. HCl (aq)  H+ + Cl- 1M x Therefore [H3O+] = 1 Therefore, pH = -log[H3O+] = -log(1) = 0 Therefore a 1M solution of HCl has pH 0.

  18. Example 3 What is the pH of a 0.01M solution of HCl? HCl (aq)  H+ + Cl- [H3O+] = 0.01 M Therefore, pH = -log[H3O+] = -log(0.01) = 2

  19. Example 4 What is the pH of a 1M NaOH solution? pOH = -log[OH-] = -log(1) = 0 pH + pOH = 14 pH = 14 – pOH pH = 14 Therefore a 1M solution of NaOH has pH 14. The pH of a very basic solution.

  20. Example 5 Determine the pH of a 0.01M NaOH solution. pOH = -log[OH-] = -log(0.01) = 2 pH + pOH = 14 pH = 14 – pOH pH = 12 Therefore a 1M solution of NaOH has pH 12. The pH of a basic solution.

  21. Example 6 The pH reading of a solution is 10.33. What is its hydrogen ion concentration? Base ten logarithm represents an exponent log10(100) =2 102 10-pH = [H+] 10-10.33 = [H+] 4.7  10-11mol/L = [H+]

  22. Example 7 Calculate the pH of a 0.00242 M H2SO4 solution . H2SO4 2H+ + SO42- 0.00242 M 0.00484M pH = - log[H+] = -log(0.00484) = -(-2.315) = 2.32

  23. Homework • Pg 239 #1, 2 • Pg 242 #5, 7, 9, 10

  24. Acid-Base Indicators • Can determine if the solution is acidic, basic or natural using various indicators • Litmus paper, bromothyomol blue, phenolphthalein are some examples. • Depending on the indicator they change colour at varying pH levels. • Need to use various indicators to solve the pH level

  25. Chart • Back of book

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