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Acids and Bases

This guide explores the essential properties of acids and bases, outlining their definitions based on Arrhenius and Bronsted-Lowry theories. It distinguishes between strong and weak acids/bases, highlighting their ionization in water and conductivity. The guide also covers conjugate acid-base pairs, monoprotic, diprotic, and triprotic characteristics, and emphasizes the pH scale's role in measuring acidity. Practical examples illustrate pH calculations with various acid and base concentrations, alongside discussions on acid-base indicators that help determine a solution's nature.

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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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