Weak Acid and Base Equilibria Solutions

Chem Worksheet - Week 9 - Solutions Weak Acid and Base Equilibria Key Questions 1. 2. 3. Given the following concentrations of H3O+ or OH–, calculate the concentration of OH– or H3O+, and indicate whether the solution is acidic or basic. [H3O+] [OH–] Acidic or basic? 1.0 x 10–9 M 1.0 x 10–5 M basic 2.5 x 10–6 M 4.0 x 10–9 M acidic 1.2 x 10–8 M 8.3 x 10–7 M basic Complete the following table by calculating the missing entries and indicate whether the solution is acidic or basic. Be sure your answers are expressed to the proper number of significant figures. [H3O+] [OH–] pH pOH acidic or basic? 5.8 x 10–5 M 1.7 x 10–10 M 4.24 9.76 acidic 1.6 x 10–9 M 6.2 x 10–6 M 8.79 5.21 basic 5.6 x 10–12 M 1.7 x 10–3 M 11.24 2.76 basic 5.0 x 10–5 M 2.0 x 10–10 M 4.30 9.70 acidic 2.0 M 5.0 x 10–15 M –0.30 14.30 acidic (very!) Determine the concentrations of H3O+ and OH– in the following solutions of strong acids or bases in water. [Caution: Think about how the acid or base dissociates in water.] Solution [H3O+] [OH–] 2.5 x 10–3 M HCl 2.5 x 10–3 M 4.0 x 10–12 M 0.0010 M NaOH 1.0 x 10–11 M 1.0 x 10–3 M 7.5 x 10–5 M Ca(OH)2 6.7 x 10–11 M 1.5 x 10–4 M In the last case, realize that two moles of OH– are produced for every one mole of Ca(OH)2 by the dissociation reaction, Ca(OH)2(aq) ÷ Ca2+(aq) + 2 OH–(aq). 4. A 0.0100 M solution of a weak acid HA has a pH of 2.60. What is the value of Ka for the acid? [Hint: What is the actual concentration of undissociated HA, [HA], in this solution?] pH = 2. 60 Y [H3O+] = 2.5 x 10–3 M We can assume that the acid is the major source of hydronium ion and that water contributes little to the total [H3O+]. This concentration of hydronium ion, then, also represents the amount of the initial HA that is lost through hydrolysis. On this basis we have the following equilibrium concentrations: Initial Equilibrium 5. HA + 0.0100 0.0100 – 0.0025 = 0.0075 H2O º H3O+ 10–7 2.5 x 10–3 + A– 0 2.5 x 10–3 What is the pH of 4.0 × 10–2 M benzoic acid (C6H5CO2H = HBz), for which Ka = 6.46 × 10–5? Is it necessary to solve the quadratic equation in this case? HBz + H2O º H3O+ + Bz– The concentration of the acid and the Ka fit our criteria to avoid needing the quadratic equation. pH = 2.79 The percent dissociation of the benzoic acid, based on our calculation, is This is acceptable, so the quadratic equation was not necessary. 6. Find the concentrations of all species and pH for a 1.0 × 10–3 M HF solution. Ka = 6.7 × 10–4 . You will need to find values for [HF], [H3O+], [F–], and [OH–]. HF + H2O º H3O+ + F– Trying the quick method, we obtain This is not acceptable. Notice that CHA = 1.0 x 10–3 and Ka = 6.7 x 10–4 are within two powers of 10 of each other, suggesting that HF is appreciably dissociated. We must solve the quadratic. This means that we will assume [HF] = CHF – [H3O+]. Substituting into the Ka expression: [H3O+]2 = 6.7 × 10–7 – 6.7 × 10–4 [H3O+] [H3O+]2 + 6.7 × 10–4 [H3O+] – 6.7 × 10–7 = 0 Solving the quadratic equation and ignoring the negative root (-1.22 × 10–3) gives [H3O+] = 5.49 × 10–4 M = [F–] Y pH = 3.26 [HF] = CHF – [H3O+] = 1.0 × 10–3 – 5.49 × 10–4 = 4.51 × 10–4 M To find [OH–], use Kw: 7. What is the pH of a 0.120 M solution of NH3(aq)? For ammonia, Kb = 1.77 × 10–5. Is it necessary to solve the quadratic equation? Justify the approach you took, either way. The hydrolysis equilibrium and the equilibrium concentrations are as follows: NH3 + H2O º NH4+ + OH– 0.120 - x x x The analytical concentration of ammonia is 0.120 M, which is greater than Kb = 1.77 × 10–5 by much more than two powers of 10. Therefore, we can probably ignore the loss of NH3 from hydrolysis. Thus, we can write and From this result, the percent hydrolysis is much less than 5%: Therefore, we are justified in ignoring loss of NH3 from hydrolysis and avoiding the quadratic equation. To obtain pH, we could calculate [H3O+] = Kw/[OH–], and then get pH; or we could calculate pOH from our value of [OH–], and get pH from pH = 14.000 – pOH. Taking the second approach, we have pOH = –log(1.48 × 10–3) = 2.830 8. Y pH = 14.000 – 2.830 = 11.170 Sodium hypochlorite, NaOCl, is the active ingredient in chlorine bleach (e.g., Chlorox©). For hypochlorous acid, HOCl, Ka = 3.0 × 10–8. a. Write the equation for the base hydrolysis equilibrium of the hypochlorite ion. OCl– + H2O º HOCl + OH– b. What is the value of Kb for the hypochlorite ion? c. What is the pH of a 0.10 M solution of sodium hypochlorite? In solution, the 0.10 M NaOCl will break up into 0.10 M Na+ and 0.10 M OCl–. Therefore, the analytical concentration of the base is . NaOCl ÷ Na+ + OCl– OCl– + H2O º HOCl + OH– 0.10 – x x x Because the analytical concentration of the base is so much greater than Kb, loss of OCl– is negligible, and we can assume Therefore, From this, pOH = 3.74, and pH = 14.00 – 4.24 = 10.26. 9. Trimethylamine, (CH3)3N, is a weak base (Kb = 6.4 × 10–5) that hydrolyzes by the following equilibrium: (CH3)3N + H2O º (CH3)3NH+ + OH– a. Write the equation for the acid hydrolysis equilibrium of trimethylammonium ion, (CH3)3NH+. (CH3)3NH+ + H2O º (CH3)3N + H3O+ b. What is the value of Ka for the trimethylammonium ion, (CH3)3NH+? . c. What is the pH of a 0.12 M solution of trimethylammonium chloride, ((CH3)3NH)Cl? ((CH3)3NH)Cl Y (CH3)3NH+ + Cl– (CH3)3NH+ + H2O º (CH3)3N + H3O+ 0.12 - x x x With Ka = 1.56 × 10–10, (CH3)3NH+ is a very weak acid, so we can definitely ignore hydrolysis and assume [(CH3)3NH+] = . pH = 5.36 10. Explain the following observations: a. H2Se is a stronger acid than H2S Se is larger than S, so the Se–H bond is weaker, making H2Se a stronger acid. b. HNO2 is a weak acid, but HNO3 is a strong acid HNO2 has one terminal oxygen, but HNO3 has two, making it a stronger acid. c. H3PO4 is a stronger acid than H3AsO4 P is more electronegative than As, making H3PO4 a stronger acid. d. FCH2CO2H is a stronger acid than CH3CO2H Electronegative fluorine withdraws electron density from the -OH bond of the carboxyl group, -CO2H, making FCH2CO2H a stronger acid.