1. 1 Acids and Bases Chapter 15 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2 Acids Have a sour taste. Vinegar owes its taste to acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas. Have a bitter taste. Feel slippery. Many soaps contain bases. Bases

3. 3 A Brønsted acid is a proton donor A Brønsted base is a proton acceptor acidbaseacidbase acid conjugate base base conjugate acid

4. Example 4 15.1 Identify the conjugate acid-base pairs in the reaction between ammonia and hydrofluoric acid in aqueous solution NH 3 (aq) + HF(aq) (aq) + F - (aq)

5. Example 5 15.1 Strategy Remember that a conjugate base always has one fewer H atom and one more negative charge (or one fewer positive charge) than the formula of the corresponding acid. Solution NH 3 has one fewer H atom and one fewer positive charge than. F - has one fewer H atom and one more negative charge than HF. Therefore, the conjugate acid-base pairs are (1) and NH 3 and (2) HF and F -.

6. 6 O H H+ O H H O H HH O H - + [] + Acid-Base Properties of Water H 2 O (l) H + (aq) + OH - (aq) H 2 O + H 2 O H 3 O + + OH - acid conjugate base base conjugate acid autoionization of water

7. 7 H 2 O (l) H + (aq) + OH - (aq) The Ion Product of Water K c = [H + ][OH - ] [H 2 O] [H 2 O] = constant K c [H 2 O] = K w = [H + ][OH - ] The ion-product constant (K w ) is the product of the molar concentrations of H + and OH - ions at a particular temperature. At 25 0 C K w = [H + ][OH - ] = 1.0 x 10 - 14 [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic

8. 8 why the value of K w for water equals 1× 10  14 At 25°C, water will dissociate very slightly into H+ and OH- ions, but the fraction is very small, only about 1 molecule in 10 million (10  7 ) will be dissociated. The other 9,999,999 molecules remain as molecules and don't dissociate. The reaction equation is: K w = [H + ][OH - ] K w = [10  7 ][10  7 ] K w = 1× 10  14 H 2 O (l) H + (aq) + OH - (aq)

9. Example 9 15.2 The concentration of OH - ions in a certain household ammonia cleaning solution is 0.0025 M. Calculate the concentration of H + ions.

10. Example 10 Strategy We are given the concentration of the OH - ions and asked to calculate [H + ]. The relationship between [H + ] and [OH - ] in water or an aqueous solution is given by the ion-product of water, K w [Equation (15.3)]. 15.2

11. Example 11 15.2 Solution Rearranging Equation (15.3), we write Check Because [H + ] < [OH - ], the solution is basic, as we would expect from the earlier discussion of the reaction of ammonia with water.

12. 12 pH – A Measure of Acidity pH = - log [H + ] [H + ] = [OH - ] [H + ] > [OH - ] [H + ] < [OH - ] Solution Is neutral acidic basic [H + ] = 1.0 x 10 -7 [H + ] > 1.0 x 10 -7 [H + ] < 1.0 x 10 -7 pH = 7 pH < 7 pH > 7 At 25 0 C pH[H + ]

13. 13 pOH = -log [OH - ] [H + ][OH - ] = K w = 1.0 x 10 -14 -log [H + ] – log [OH - ] = 14.00 pH + pOH = 14.00 Other important relationships pH Meter

14. Example 14 15.3 The concentration of H + ions in a bottle of table wine was 3.2 x 10 -4 M right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to 1.0 x 10 -3 M. Calculate the pH of the wine on these two occasions.

15. Example 15 15.3 Strategy We are given the H + ion concentration and asked to calculate the pH of the solution. What is the definition of pH?

16. Example 16 15.3 Solution According to Equation (15.4), pH = -log [H + ]. When the bottle was first opened, [H + ] = 3.2 x 10 -4 M, which we substitute in Equation (15.4) pH = -log [H + ] = -log (3.2 x 10 -4 ) = 3.49 On the second occasion, [H + ] = 1.0 x 10 -3 M, so that pH = -log (1.0 x 10 -3 ) = 3.00

17. Example 17 15.3 Comment The increase in hydrogen ion concentration (or decrease in pH) is largely the result of the conversion of some of the alcohol (ethanol) to acetic acid, a reaction that takes place in the presence of molecular oxygen.

18. Example 18 15.4 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. Calculate the H + ion concentration of the rainwater.

19. Example 19 15.4 Strategy Here we are given the pH of a solution and asked to calculate [H + ]. Because pH is defined as pH = -log [H + ], we can solve for [H + ] by taking the antilog of the pH; that is, [H + ] = 10 -pH, as shown in Equation (15.5).

20. Example 20 15.4 Solution From Equation (15.4) pH = -log [H + ] = 4.82 Therefore, log [H + ] = -4.82 To calculate [H + ], we need to take the antilog of -4.82 [H + ] = 10 -4.82 = 1.5 x 10 -5 M

21. Example 21 15.4 Check Because the pH is between 4 and 5, we can expect [H + ] to be between 1 x 10 -4 M and 1 x 10 -5 M. Therefore, the answer is reasonable.

22. Example 22 15.5 In a NaOH solution [OH - ] is 2.9 x 10 -4 M. Calculate the pH of the solution.

23. Example 23 15.5 Strategy Solving this problem takes two steps. First, we need to calculate pOH using Equation (15.7). Next, we use Equation (15.9) to calculate the pH of the solution. Solution We use Equation (15.7): pOH = -log [OH - ] = -log (2.9 x 10 -4 ) = 3.54

24. Example 24 15.5 Now we use Equation (15.9): pH + pOH = 14.00 pH = 14.00 - pOH = 14.00 - 3.54 = 10.46 Alternatively, we can use the ion-product constant of water, K w = [H + ][OH - ] to calculate [H + ], and then we can calculate the pH from the [H + ]. Try it. Check The answer shows that the solution is basic (pH > 7), which is consistent with a NaOH solution.

25. Example 25 Describe two methods of measuring pH 1)Indicators: could be red or blue litmus paper ( red becomes blue with base, and blue becomes red with acid) Or could be dyes like methyl orange 2)pH meter: A pH meter, also called a pH tester, measures the acidity and alkalinity of a liquid or other substance, on a scale with a range of 0 to 14 where pH = 7 is neutral pH > 7 is alkaline pH < 7 is acidic SWQ