pH and pOH Calculations

KEY Chemistry: pH and pOH calculations Part 1: Fill in the missing information in the table below. pH [ H3O1+ ] pOH [ OH1– ] ACID or BASE? 3.78 1.66 x 10–4 M 10.22 6.03 x 10–11 M Acid 3.41 3.89 x 10–4 M 10.59 2.57 x 10–11 M Acid 8.81 1.55 x 10–9 M 5.19 6.46 x 10–6 M Base 8.69 2.04 x 10–9 M 5.31 4.88 x 10–6 M Base 8.46 3.47 x 10–9 M 5.54 2.88 x 10–6 M Base 12.1 8.45 x 10–13 M 1.90 1.26 x 10–2 M Base 11.86 1.38 x 10–12 M 2.14 7.24 x 10–3 M Base 3.40 3.98 x 10–4 M 10.6 2.31 x 10–11 M Acid 10.91 1.23 x 10–11 M 3.09 8.13 x 10–4 M Base 5.13 7.49 x 10–6 M 8.87 1.35 x 10–9 M Acid 4.06 8.71 x 10–5 M 9.94 1.15 x 10–10 M Acid 6.41 3.89 x 10–7 M 7.59 2.57 x 10–8 M Acid 4.16 6.92 x 10–5 M 9.84 1.45 x 10–10 M Acid 0.98 1.06 x 10–1 M 13.0 1.00 x 10–13 M Acid 10.18 6.61 x 10–11 M 3.82 1.51 x 10–4 M Base 7.93 1.17 x 10–8 M 6.07 8.53 x 10–7 M Base 7.05 8.91 x 10–8 M 6.95 1.12 x 10–7 M ~Base 9.33 4.73 x 10–10 M 4.67 2.14 x 10–5 M Base 12.67 2.14 x 10–13 M 1.33 4.68 x 10–2 M Base 12.0 1.0 x 10–12 M 2.01 9.87 x 10–3 M Base 11.68 2.09 x 10–12 M 2.32 4.79 x 10–3 M Base 7.04 9.22 x 10–8 M 6.96 1.10 x 10–7 M ~Base 1.76 1.74 x 10–2 M 12.24 5.75 x 10–13 M Acid 2.70 2.00 x 10–3 M 11.3 5.39 x 10–12 M Acid Part 2: For each of the problems below, assume 100% dissociation. 1. A. Write the equation for the dissociation of hydrochloric acid. HCl(aq) → H1+(aq) + Cl1-(aq) B. Find the pH of a 0.00476 M hydrochloric acid solution. HCl(aq) → H1+(aq) + Cl1-(aq) 0.00476 M 0.00476 M pH = - log [H+ ] pH = - log [0.00476 M] pH = 2.32 2. A. Write the equation for the dissociation of sulfuric acid. H2SO4(aq) → 2 H1+(aq) + SO42-(aq) B. Find the pH of a solution that contains 3.25 g of H2SO4 dissolved in 2.75 liters of solution.  1 mol H 2 SO 4   = 0.033 mol H 2 SO 4 Step 1 : x mol H 2 SO 4 = 3.25 g H 2 SO 4   98 g H 2 SO 4  0.033 mol H 2 SO 4 mol Step 2 : M =  M=  M = 0.0121M H 2 SO 4 L 2.75 L Step 3 : H 2 SO 4 (aq ) → 2 H+ + SO 4 2− (aq )  0.0242M 0.0121M + Step 4 : pH = - log [H ]  pH = - log [0.0242 M]  pH = 1.62 3. A. Write the equation for the dissociation of sodium hydroxide. NaOH(aq) → Na1+(aq) 0.000841 M B. + OH1-(aq) 0.000841 M Find the pH of a 0.000841 M solution of sodium hydroxide. pOH = - log [OH- ] pOH = - log [0.000841M] pOH = 3.08 pH + pOH = 14 pH + 3.08 = 14 pH = 10.92 or K w = [H+ ][OH- ] pH = - log[H + ] 1 10 -14 = [H+ ][0.000841M] pH = - log [1.19  10 -11 M] pH = 10.92 [H+ ] = [1.19  10 -11 M] Part 2: continued 4. A. Write the equation for the dissociation of aluminum hydroxide. Al(OH)3(aq) → Al3+(aq) 2.36x10-5 M B. + 3 OH1-(aq) 7.08x10-5 M If the pH is 9.85, what is the concentration of the aluminum hydroxide solution? pH + pOH = 14 9.85 + pOH = 14 pOH = 4.15 pOH = - log [OH- ] 7.08 x 10 -5 M = 2.36 x 10 - 5 M 3 4.15 = - log [OH- ] 2nd log - 4.15 = [OH- ] [OH- ] = 7.08  10-5 M 5. A. Write the equation for the dissociation of calcium hydroxide. Ca(OH)2(aq) → Ca2+(aq) 2.18x10-3 M B. + 2 OH1-(aq) 4.37x10-5 M If the pH is 11.64 and you have 2.55 L of solution, how many grams of calcium hydroxide are in the solution? pH + pOH = 14 11.64 + pOH = 14 pOH = 2.36   pOH = - log OH- 2.36 = - log [OH- ] 2.18  10 -3 M 2 4.37  10 − 3 M 2nd log - 2.36 = [OH- ] [OH- ] = 4.37  10-3 M M= x mol Ca(OH) 2 mol  2.18  10 -3 M   x = 5.57  10 -3 mol Ca(OH) 2 L 2.55 L  74 g Ca(OH) 2 x g Ca(OH) 2 = 5.57  10 -3 mol  1 mol Ca(OH) 2   = 0.412 g Ca(OH) 2  1a. HCl(aq) → H1+(aq) + Cl1-(aq) pH and pOH b. HCl(aq) → H1+(aq) + Cl1-(aq) 0.00476 M 0.00476 M pH = - log [H+ ] pH = - log [0.00476 M] pH = 2.32  1 mol H2 SO 4   = 0.03 mol H2 SO 4 Step 1 : x mol H2SO 4 = 325 g H2 SO 4   98 g H2 SO 4  0.03 mol H2SO 4 mol Step 2 : M =  M=  M = 0.01M H2 SO 4 L 2.75 L 2a. Step 3 : H2SO 4 (aq ) → 2 H+ + SO 4 2− (aq )  0.02 M 0.01M Step 4 : pH = - log [H+ ]  pH = - log [0.02 M]  pH = 1.70 3. NaOH(aq) → Na1+(aq) 0.000841 M + OH1-(aq) 0.000841 M pOH = - log [OH- ] pOH = - log [0.000841M] pOH = 3.08   pH = - log1.19  10 pH = - log H+ -11 pH + pOH = 14 pH + 3.08 = 14 pH = 10.92 K w = [H+ ][OH- ] or 1 10 -14 = [H+ ][0.000841M] [H+ ] = [1.19  10 -11 M]  M pH = 10.92 4. Al(OH)3(aq) → Al3+(aq) 2.36x10-5 M + 3 OH1-(aq) 7.08x10-5 M pH + pOH = 14 9.85 + pOH = 14 pOH = 4.15 pOH = - log [OH- ] 4.15 = - log [OH- ] 7.08 x 10 -5 M = 2.36 x 10 - 5 M 3 2nd log - 4.15 = [OH- ] [OH- ] = 7.08  10-5 M 5. Ca(OH)2(aq) → Ca2+(aq) 2.18x10-3 M pH + pOH = 14 11.64 + pOH = 14 pOH = 2.36 + 2 OH1-(aq) 4.37x10-5 M   pOH = - log OH- 2.36 = - log [OH- ] 2.18  10 -3 M 2 4.37  10 − 3 M 2nd log - 2.36 = [OH- ] [OH- ] = 4.37  10-3 M x mol Ca(OH) 2 mol  2.18  10 - 3 M   x = 5.57  10 - 3 mol Ca(OH) 2 L L  74 g Ca(OH) 2   = 0.412 g Ca(OH) 2 x g Ca(OH) 2 = 5.57  10 - 3 M  1 mol Ca(OH) 2  M=