Acids - Base Equilibria Part I: pH and Acidity

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Presentation transcript:

Acids - Base Equilibria Part I: pH and Acidity Jespersen Chap. 17 Sec 1 & 2 Dr. C. Yau Spring 2013 1 1

Autoionization of Water H2O + H2O H3O+ + OH 2 H2O H3O+ + OH With H2O as a solvent, [H2O] becomes a constant. We rearrange the equation to give Kc [H2O]2 = [H3O+][OH] Kw = [H3O+][OH] For simplicity, we write Kw = [H+][OH] At 25°C, Kw = 1.0x10-14 = [H+][OH] Know well! Note: 2 sig. fig. for 1.0 2 2

Kw = 1.0x10-14 = [H+][OH] As usual, changes in concentration does not change Kw. They only cause shifts in the equilibrium. Only T can change Kw. Table 17.1 p. 775

Kw = 1.0x10-14 = [H+][OH] Neutral solutions: [H+]=[OH] Acidic solutions: [H+]>[OH] Basic solutions: [H+]<[OH] Note how small Kw is. This means water is a VERY weak acid. This is why it is considered nonelectrolytic. It does not conduct electricity unless it is contaminated with ionic compounds. Tap water IS electrolytic because it is not pure water and contains salts.

H2O auto-ionization occurs in any solution When other ions present [H+]  [OH] But Kw = [H+]·[OH] = 1.0 x 1014 In aqueous solution, Product of [H+] and [OH] equals Kw [H+] and [OH] may not actually equal each other

(very slightly basic, more OH- than H+) In a sample of blood at 25°C, [H+] is 4.6x10-8 M. Find the molar concentration of OH-, and decide if the sample is acidic, basic, or neutral. [H+][OH]= 1.0x10-14 Compare Ans. [OH-] = 2.2x10-7 M OH- (very slightly basic, more OH- than H+) Pract Exer 17.1 & 17.2 p. 776

pH Scale When we have a dilute solution of acid, the H+ concentration is very small, such as 0.0001M, and it becomes cumbersome to be talking about H+ concentration of 0.0001 M or 1x10-4M. The pH scale was invented to make it easier to communicate the acidity of a solution. KNOW THIS WELL: pH = - log [H+] or pH = - log [H3O+]

e.g. [H+] = 2.43x104 M, what is the pH? pH =  log [H+] =  log 2.43x104 (3 sig.fig.) =  (3.614) (3 decimal places) = 3.614 e.g. pH = 8.71, what is the [H+]? 8.71=  log [H+] 8.71 = log [H+] Antilog(-8.71) = antilog log [H+] 10-8.71 = [H+] (2 decimal places) Ans. [H+] = 1.9x10-9 M (2 sig. fig.) REMEMBER: 10-pH = [H+]

We see that the pH and H+ concentration has an inverse relationship: As H+ concentration increased, the pH decreases. The smaller the pH, the more acidic it is. The larger the pH, the more basic it is. So, if the swimming pool water has a pH that is too high, that means it is too basic.

e.g. If [OH] = 4.2x103M, what is the pOH? pOH =  log [OH-] e.g. If [OH] = 4.2x103M, what is the pOH? pOH =  log 4.2x103 (2 sig.fig.) =  2.38 = +2.38 (2 decimal places) e.g. If pOH = 12.3, what is the hydroxide conc? [OH] = 1012.3 (1 decimal place) = 5x1013 M (1 sig. fig.) THIS IS HOW pH AND pOH ARE RELATED: Remember that [H+] [OH] = 1.0x1014 Finding the log of both sides of the equation gives us… log [H+] + log [OH] = log 1.0x1014 Multiply both sides of equation by 1. ( log [H+] + log [OH]) = ( log 1.0x1014) ( log [H+] )+(  log [OH]) =  (14.00) pH + pOH = 14.00 REMEMBER THIS!

Practice Exer. 17.4 p. 778 Because rain washes pollutants out of the air, the lakes in many parts of the world have undergone pH changes. In a New England state, the water in one lake was found to have a hydrogen ion conc of 3.2x10-5 mol L-1. What are the calculated pH and pOH? Is it acidic or basic? Ans. pH = 4.49 (acidic) pOH = 9.50

What is the pH of a NaOH soln at 25 oC in which the hydroxide ion concentration equals 0.0026 M? Ans. pH = 11.41 (basic, as expected). Practice Exercises 17.3 & 17.5 p. 778

Calculate the values of pH, pOH and [OH-] for the following solutions: a) 0.020M HCl b) 0.00035 M Ba(OH)2. You must first note that HCl is a strong acid. Only because this is so can you assume 0.020M HCl means 0.020 M H+. Ans. a) pH = 1.70, pOH = 12.30, [OH-]=5.0x10-13M b) 0.00035M Ba(OH)2 means [OH-] = ? Ans. [OH-] = 0.00070M, pOH = 3.15, pH = 10.85

Acid or basic solutes suppress the ionization of water. H2O + H2O H3O+ + OH 2 H2O H3O+ + OH Remember Le Chatelier’s Principle: If an acidic solute (e.g. HCl) is added to water, it shifts the equilibrium to the left, thus suppressing the ionization of water. This means that in an acidic solution the only source of H+ is from the acid and essentially no contribution from water. In the same way, in a basic solution the only source of OH- is from the base, none from the dissociation of water.

Calculating without a calculator Often we do not need a calculator to calculate pH or [H+]. [H+] = 108 What is the pH? [OH] = 106 What is the pH? pH = 4 What is the pOH? pOH = 3 What is the pH?

A soln was made by dissolving 0. 837 g Ba(OH)2 in 100 mL final volume A soln was made by dissolving 0.837 g Ba(OH)2 in 100 mL final volume. What is the molar conc of OH- in the soln? What are the pOH and the pH? What is the hydrogen ion concentration in the solution? MM(Ba(OH)2) = 171.3 g mol-1 Ans. [OH-] = 0.0977 M, pOH = 1.010 pH = 12.990, [H+] = 1.02x10-13M Practice Exer 6, 7, & 8 p. 779

What is the pH of 0.00000001 M HCl? [H+] =1x108M pH = 8 (basic) How can a solution of HCl be basic?? Answer: Its pH will NOT be 8. At very dilute solutions, the H+ will NOT suppress the dissociation of water. Water will contribute to the H+ concentration and keep it from becoming basic! Ans. pH = 7 (stays neutral)