Chapter 15 Acids and Bases

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) H2O

2. Arrhenius also proposed a useful definition for bases. Bases - produce OH⁻ ions in aqueous solution. NaOH(s) Na⁺(aq) + OH⁻ (aq) H2O

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 occursHA (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) • Acid base conj. acid conj. Base • H₂SO₄(aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HSO₄⁻ (aq) • Acid Base conj. Acid conj. Base • H₂S(aq) + H₂O₍ι₎ → H₃O⁺ (aq) + HS⁻ (aq) • Acid base conj. acid conj. 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₎ • Acid base salt water

13. Other salts are possible H₂CO₃ + Mg(OH)₂ → MgCO₃ + HOH Acid base salt 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 metal aqueous acid salt hydrogen gas Mg + H₂SO₄ → MgSO₄ + H₂ Solid metal aqueous acid salt 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