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17 Additional Aspects of Solutions to Exercises Aqueous Equilibria 5.56 X 10⁻¹⁰ = 0.01667 ;x = = 3.043 X 10⁻⁶ = 3.0 X 10⁻⁶ M;pH=5.52 (e) Past the equivalence point, from the excess determines the pH. = 2.5 0.0910 X 10⁻⁵ L mol = 2.747 X = 2.7 10⁻⁴ M;pH=3.56 (f) Past the equivalence point, [H⁺] from the excess determines the pH. 1.25 10⁻⁴ L mol = 1.316 X 10⁻³ = 1.3 10⁻³ M;pH=2.88 0.0950 17.47 Analyze/Plan. Calculate the pH at the equivalence point for the titration of several bases with 0.200 M HBr. The volume of 0.200 M HBr required in all cases equals the volume of base and the final volume = The concentration of the salt produced at the equivalence point is 0.200 M X = 0.100 M. In each case, identify the salt present at the equivalence point, determine its acid-base properties (Section 16.9), and solve the pH problem. Solve. (a) NaOH is a strong base; the salt present at the equivalence point, NaBr, does not affect the pH of the solution. 0.100 M NaBr, pH = 7.00 (b) is a weak base, so the salt present at the equivalence point is Br-. This is the salt of a strong acid and a weak base, so it produces an acidic solution. 0.100 M 1L (aq) + [equil] Assume is small with respect to [salt]. (c) C₆H₅NH₂ is a weak base and C₆H₅NH₃⁺ is an acidic salt. 0.100 M Proceeding as in (b): 17.48 The volume of NaOH solution required in all cases is 522