Question

Given that K, for HCN is 6.2 x 10-10 at 25 °C, what is the value of Ky for CN- at 25 °C? Given that Kh for CH, NH, is 5.0 x 10-4 at 25 °C, what is the value of K, for CHNHĨ at 25 °C? Kg =

Answer 1

Solution, V is a weak monopnotic acid and it's dissociation Equilibrium bien dopuna est to a HOW H+ + CN-t ka = 6.2x10- The Equilibri Constant is given by ka= [H+] CON- -------(1). Снен) And Consider the equilibrium for it is conjugate base: (tu CN- & H2O HCN + OH- ; K6=? Nixon we The Equilibrium Constant is given by WXc K62 [HCN] [OH-] [CN-]

Multiplying the equation ③ and ④; we get- Kaxkn = [HT] [CA) CHCN COH-] - CHÓW an [ar] - [H+] [OH-] A Kaxk6= kw where Kw = [H +) [OH-] is the ionic Product of water. Kw = 10-14 at 25°C temperatene. i kaxk6= 10-14 g (6.2 x 10-10) x K4 = 104 er, K6 = 1.62 x 10 -5. Hence, The value Kb = 1.62 x 10-5 for cw at 25°c. a 25. 2. The The Protonation of the base and it's Equilibrium: CH₂ Mt 2 + H₂O CH₂NH₂ + OH- Kb = 5.0*10-4. The Equilibrium Constent is given by - Ссыз кі, )Сон-) ----() [Chanta] And the bandrolongation of the Conjugate acid and it's Equilibrim. CH₃nt - CH₂ Mt₂ + H+ kaan The Equilibrium Constant is given by india [CH₂ MHz] CHT) [ @Higitale) -----(2) multiplying the equation ① and @ we get- kkx kа з сэнэЈСен сенз 2 )Сн+) Сенутна) * Снут, ) I = [OH-][ht] w KbXka = kw. a, 5.0*10-4 X ka = 10-14 kW = 10-14 at 25°c) Ionic Product of water. er, ka = 2 x 10-11 Hence, the value ka= 2.0x10-11 for CH₂ mtz at 28"c.