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17 Additional Aspects of Solutions to Exercises Aqueous Equilibria [H⁺] 1.8 10⁻⁵ (0.07446+x) (1.079 ≈ 2.6 10⁻⁷ M (d) At the equivalence point, only is present. 0.0700 L The pertinent equilibrium is the base hydrolysis of + 1L CH₃COOH(aq) + initial 0.0750 M 0 0 equil 5.56 X 10⁻¹⁰ = 0.0750 ; ≈ 5.56 X 10⁻¹⁰ (0.0750); = 6.458 10⁻⁶ = 6.5 10⁻⁶ M (e) After the equivalence point, the excess strong base determines the pOH and pH. The [OH⁻] from the hydrolysis of is small and can be ignored. 0.075 0.0705 =1.064 X 10⁻³ = 1.1 10⁻³ M; pOH = 2.97 (f) = 0.0265 0.0850 17.46 (a) Weak base At equilibrium, = = = 1.8 10⁻⁵ = ≈ = = 10⁻⁴ = 9.5 10⁻⁴ M (b-f) Calculate mol NH₃ and mol after the acid-base reaction takes place. 0.050 M NH₃ X 0.0300 L = 1.5 X mol NH₃ present initially. 520