Equilíbrio Iônico Aquoso

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370 Chapter 17 Aqueous Ionic Equilibrium 17.77 Given: 0.229 g unknown monoprotic acid titrated with 0.112 M NaOH and curve Find: molar mass and pKₐ of acid Conceptual Plan: The equivalence point is where sharp pH rise occurs. The is the pH at a volume of added base equal to half the equivalence point volume. Then mL NaOH L NaOH 1L 1000 mL then [NaOH], L NaOH mol NaOH = mol acid then mol acid, g acid molar mass. g acid M L mol acid Solution: The equivalence point is at 25 mL NaOH. The pH at 0.5 X 25 mL = 13 mL is ~3 = pKₐ. Then 25 NaOH X 1000 1L = 0.025 L NaOH then 0.112 mol NaOH X 0.025 L = 0.0028 mol NaOH = 0.0028 mol acid then molar mass = 0.0028 0.229 mol g acid acid = 82 g/mol Check: The units (none and g/mol) are correct. The pKₐ is consistent with a weak acid. The molar mass is reasonable for an acid (> 1 g/mol). 17.79 Given: 20.0 mL of 0.115 M sulfurous acid (H₂SO₃) titrated with 0.1014 M KOH Find: volume of base added Conceptual Plan: Because this is a diprotic acid, each proton is titrated sequentially. Write balanced equations. H₂SO₃ + + and + + Then mL L then mol then set mol base = mol acid (KOH) and 1L M balanced equation has 1:1 stoichiometry (1st equivalence point) 1000 mL L [KOH], mol L KOH mL KOH the volume to the second equivalence point will be 1000 mL M L 1L double the volume to the first equivalence point. Solution: 20.0 X 1000 1L mL = 0.0200 L then 0.115 mol H₂SO₃ X 0.0200 L = 0.00230 mol So mol base = mol H₂SO₃ = 0.00230 mol = mol KOH then 0.00230 X 0.1014 1L = 0.02268245 LKOH X 1000 = 22.7 mL KOH to first equivalence point. The volume to the second equivalence point is simply twice this amount, or 45.4 mL, to the second equivalence point. Check: The units (mL) are correct. The volume of base is greater than the volume of acid because the concentration of the acid is a little greater that of the base. The volume to the second equivalence point is twice the volume to the first equivalence point. 17.81 The indicator will be in its acid form at an acidic pH, so the color in the sample will be red. The color change will occur over the pH range from pH = pKₐ - 1.0 to pH = pKₐ + 1.0, so the color will start to change at pH = 5.0 1.0 = 4.0 and finish changing by pH = 5.0 + 1.0 = 6.0. 17.83 Because the exact conditions of the titration are not given, a rough calculation will suffice. Recall that at the equiva- lence point, the moles of acid and base are equal. If it is assumed that the concentrations of the acid and the base are equal, the total volume of the solution will have doubled. Assuming an acid and base concentration of 0.1 M, the conjugate based formed must have a concentration of ~0.05 M. From earlier calculations, it can be seen that the Kb = Kw = 0.05 thus, = 0.05 = 5 and the pH = 14 log Copyright © 2017 Pearson Education, Inc.