1 (513)

Prévia do material em texto

17 Additional Aspects of Solutions to Exercises Aqueous Equilibria = 1.8 10⁻⁵ = [CH₃COOH] = (x) L ≈ x(0.010) 0.0075 = 1.35 10⁻⁵ = 1.4 pH = 4.87 Check. for = 4.74. [CH₃COO⁻] > [CH₃COOH], pH of buffer = 4.87, greater than 4.74. 17.16 Analyze/Plan. Follow the logic in Sample Exercise 17.1. Solve. (a) HCOOH is a weak acid, and HCOONa contains the common ion the conjugate base of HCOOH. Solve the common-ion equilibrium problem. HCOOH(aq) 1L + i 0.100 M 0.250 M -x +x +x e (0.100 x) M +x M (0.250 + x) M X = 7.20 = 10⁻⁵ M = pH = 4.14 Check. Since the extent of ionization of a weak acid or base is suppressed by the presence of a conjugate salt, the 5% rule usually holds true in buffer solutions. (b) C₅H₅N is a weak base, and contains the common ion C₅H₅NH⁺, which is the conjugate acid of C₅H₅N. Solve the common ion equilibrium problem. C₅H₅N(aq) + 1L + (aq) i 0.510 M 0.450 M -x +x +x e (0.510 x) M (0.450 + M +x M Kb = 1.7 10⁻⁹ = [C₅H₅N] = (0.450 (0.510-x) + (x) ≈ 0.450 0.510 = 1.927 = 1.9 10⁻⁹ M = pOH = 8.715, pH = 14.00 - 8.715 = 5.29 Check. In a buffer, if [conj. acid] > [conj. base], pH of the conj. acid. In this buffer, of is 5.23. [C₅H₅NH⁺]