1. Chapter 15 Acids and Bases First defined by Svante Arrhenius 1903 Nobel Prize winner who proposed that: Acids - produce H⁺ ions in aqueous solution. HCl (g) H⁺ (aq) + Cl ⁻ (aq) H2OH2O

2. Arrhenius also proposed a useful definition for bases. Bases - produce OH⁻ ions in aqueous solution. NaOH (s) Na⁺ (aq) + OH ⁻ (aq) H2OH2O

3. Acids and bases using according to the Bronsted-Lowry model. Acid - proton donor Base – proton acceptor According to this model water, due to its polar nature, aids in removing the proton.

4. In solution, the following reaction occurs HA (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + A⁻ (aq) In this example: HA donates a proton to water which would make HA the acid and water accepts the proton which make it a base. The conjugate acid is H₃O⁺ the hydronium ion. The conjugate base is A⁻, which is everything that remains after the acid donates its proton

5. Using a specific acid HBr (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + Br⁻ (aq) Acidbaseconj. acidconj. Base H₂SO₄ (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HSO₄⁻ (aq) AcidBaseconj. Acidconj. Base H₂S (aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HS⁻ (aq) Acidbaseconj. acidconj. Base With Bronsted-Lowry acid base theory some themes persist. Water is the base, the conjugate acid is H₃O⁺ and the conjugate base is simply the acid without a proton (H⁺ ion)

6. Differences between the 2 theories With Arrhenius, acidic aqueous solutions contain the H⁺ ion. With Bronsted-Lowry H₂O is included in the product, so a Hydronium ion (H₃O⁺) is found in acidic solutions.

7. The Hydronium ion, H₃O⁺ In water, H⁺ combines with water to form the hydronium ion H⁺ + H₂O → H₃O⁺ For our purposes H⁺ and H₃O⁺ can be used interchangeably.

8. Including H₂O as a reactant The ionization of an acid would look slightly different if water is included as a reactant. Let’s use HBr as an example HBr + H₂O → H₃O⁺ + Br⁻ Presence of either H⁺ or H₃O⁺ make a solution acidic the difference is only how we choose to write the ionization.

9. Water is amphoteric Amphoteric - a substance that can behave as either an acid or a base. Equation for the ionization of water H₂0₍ι₎ + H₂0₍ι₎ → H₃O⁺(acid) + OH⁻(base) Or H₂O → H⁺(acid) + OH⁻(base)

10. [ # ] Placing a number in brackets means that that number then represents the a concentration in “moles per liter” aka Molarity [OH⁻] = 1.6 Means that the hydroxide ion concentration is 1.6 M

11. Comparing [H⁺] to [OH⁻] In neutral solution [H⁺] = [OH⁻] In acidic solution [H⁺] > [OH⁻] In basic solution [H⁺] < [OH⁻]

12. Neutralization Reaction This is usually a double replacement reaction with an acid as a reactant and a base as the other reactant. The products are salt and water. Ex. HCl (aq) + NaOH (aq) → NaCl (aq) + HOH₍ι₎ Acidbase saltwater

13. Other salts are possible H₂CO₃ + Mg(OH)₂ → MgCO₃ + HOH Acidbasesalt water HNO₃ + LiOH → LiNO₃ + HOH Acid base salt water

14. The pH scale This is a convenient way to express relatively small [H⁺] Generally, pH results should range from 0-14. To calculate pH you will use the log function on your calculator.

15. pH values of common materials

16. Calculating pH pH = - log [H⁺] Remember, [H⁺] is the hydrogen ion concentration expressed in mol/L a.k.a. MOLARITY

17. Calculating pH An aqueous solution has a [H⁺] =2.97x10⁻⁶ what is the pH of the solution? pH = - log [H⁺] pH = - log [2.97x10⁻⁶] 3 sig figs pH = 5.527 3 sig figs

18. Calculating [H⁺] from pH This operation involves the inv log function or the antilog so we will have an activity sheet designed to help you to enter this correctly in your individual calculator. A solution has a pH of 4.58 what is the [H⁺]? [H⁺] = 2.6 x 10⁻⁵ M

19. An HCl solution has a [H⁺] of 1.4 x 10⁻⁴ M, what is the pH? pH = - log [H⁺] pH = - log [1.4 x 10⁻⁴] pH = 3.85

20. Calculating pOH The calculation is identical to the pH calculation except it involves the [OH⁻]. pOH = - log [OH⁻]

21. Calculating pOH What is the pOH for a solution whose [OH⁻] = 1.89 x10⁻⁶ ? pOH = - log [OH⁻] pOH = - log [1.89 x 10⁻⁶] pOH = 5.724

22. What is the pOH of a solution with an [OH⁻] = 3.33 x 10⁻⁴ ? pOH = - log [OH⁻] pOH = - log 3. 33 x 10⁻⁴ pOH = 3.478

23. Calculating [OH⁺] from pH This operation involves the inv log function or the antilog so we will have an activity sheet designed to help you to enter this correctly in your individual calculator. A solution has a pOH of 5.55 what is the [OH⁺]? [OH⁺] = 2.8 x 10⁻⁶ M

24. Calculating: pOH from pH, pOH from pH Due to experiment/observation we know that pH + pOH = 14.00 So we can convert between pH + pOH with a simple subtraction problem.

25. pH and pOH conversions The pH of a solution is 2.29, what is the pOH? pOH = 14.00 – pH pOH = 14.00 – 2.29 pOH = 11.71 The pOH of a solution is 6.65, what is the pH? pH = 14.00 – pOH pH = 14.00 - 6.65 pH = 7.35

26. Reactions of Acids with Metals Many acids will react with the more reactive metals to produce the salt of that acid and hydrogen gas. Ex. 2Al + 6HCl → 2AlCl₃ + 3H₂ Solid metalaqueous acidsalthydrogen gas Mg + H₂SO₄ → MgSO₄ + H₂ Solid metalaqueous acidsalt hydrogen gas

27. Acidity of Basicity on pH Scale Summary: pH = 7 Neutral pH < 7 Acidic pH > 7 Basic

28. Ion Product Constant In aqueous solution [OH⁻] [H⁺] = 1.00 x 10⁻¹⁴

29. Ex. If an aqueous solution has a [H⁺] = 1.83 x 10⁻⁵ what is the [OH⁻] ? [OH⁻] = 1.00 x 10⁻¹⁴ / 1.83 x 10⁻⁵ [OH⁻] = 5.46 x 10⁻¹⁰

30. If an aqueous solution has an [OH⁻] of 5.43 x 10⁻⁴, what is the [H⁺] ? [H⁺] = 1.00 x 10⁻¹⁴ / [OH⁻] [H⁺] = 1.00 x 10⁻¹⁴ / 5.43 x 10⁻⁴ [H⁺] = 1.84 x 10⁻¹¹