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17 Additional Aspects of Solutions to Exercises Aqueous Equilibria 0.15 M X 0.0500 mL = 7.5 X 10⁻³ mol 0.15 M 0.0500 mL = 7.5 X 10⁻³ mol HCOO⁻ = 7.5 10⁻³ mol HCOOH Solve the weak acid problem to determine and [HCOOH] at equilibrium. = (0.075 x) M 1.8 10⁻⁴ = (0.075-x) x² ≈ 0.075 ; = 3.7 X 10⁻³ M and HCOO⁻ [HCOOH] = (0.075 0.0037) = 0.071 M [HCOOH] [H⁺] 100 = 0.075 10⁻³ 100 = 4.9% dissociation In summary: [Na⁺] = = 0.075 M, [HCOOH] = 0.071 M, = = 0.0037 M 17.109 (a) For a monoprotic acid (one per mole of acid), at the equivalence point moles OH- added = moles originally present X = g acid/molar mass MM = MB g acid X = 0.0500 M 0.1044g 0.02210 L = (b) 11.05 mL is exactly half-way to the equivalence point (22.10 mL). When half of the unknown acid is neutralized, [HA] = = and pH = (c) From Appendix D, Table D.1, acids with Ka values close to 1.3 are name Ka formula molar mass propionic 1.3 10⁻⁵ 74.1 butanoic 1.5 10⁻⁵ 88.1 acetic 1.8 CH₃COOH 60.1 hydroazoic 1.9 43.0 Of these, butanoic has the closest match for and molar mass, but the agreement is not good. 546