Question

3 attempts left Check my work Be sure to answer all parts. A volume of 25.0 mL of 0.120 M HCl is titrated against a 0.120 M CH3NH2 solution added to it from a buret. (a) Calculate the pH value of the solution after 10.0 mL of CH3NH2 solution has been added. (6) Calculate the pH value of the solution after 25.0 mL of CH NH, solution have been added. (c) Calculate the pH value of the solution after 35.0 mL of CH3NH2 solution have been added.

Answer 1

a: Rxn involved 3 H + Hugo Ĉ CHE NH +040 Kb = 4.4*10-4 cHa NH, HĐ Chy NHg OH H Aਲ ਆ । CH3NH2 + He owes Hol Kw Part (as total = 35m! = 4.4x10-4 10t4 = 4.4x10+14 . t. Crg NH3OH at Hom a Chy NH3 (limiting) •120x 10mL 0:120x25 ml - + Holice) 35mL 35 mL - 120x10 -:120X10 35 - 120X10 + 0.120X10 0.120 X 10 - 35 35 • 120 x 10 •120X15 35 120x10 کی . Strong acid is present at equilibrium, pH will be decided by that only.

pH = -log [ = -log[ 0-12.0X IS] | | PH = -2.97 1:29 ② When 85 mL of Cha Nng is added., Neutralization is done, equivalence point has reached. CH3NHgOH + H O Ž CHg NH3 H + Heures . 0.120x 25ml 50ml 0:12x25 ml ome -0.06M -0.06M 6M +0.06M - 0.06M only → CH₃NH₂O (Conjugate acid) is hydrolyses of CHg NH3 will CH₃ NH3 + Hooces left at equilibrium) take plance. . Ť CH3NH₂ OH + H I 0.06M ty J st 0.06-Y 0.06 :: Keq = Kna 10-14 Lp. 4xlo-4 = V. 973 x10-11,

- keq = kn is very small. - Value of y will be Small, as much in forward direction. reaction will not proceed Kn 4.4 0.06 = (*) CY jopa 10=10 40000 vp = 1.3636X 10-18 10-10 x0.06 4.4 [ht] = = 1.168x10–4] pH = -log [H +3 = pH = -log (1.168x10-6) 5.93 ® when 35 mL of CH3NHg is added: VTotal = 35+ 25 = Gomb. E CHE NHGOH + H = CHI NHL Base) (Conjugate + oHo acid) 0:120x25 H 0.120 x 35 60 = 0.07 0:05 -0.05 C -0.05 +0.05 +0.05 4 0.02 - Buffer solution is formed: Ki = 4.4x10~7 . pot of Baric buffer: pko = 4-10g 4.4 poh= plot log I Conjugate acid 7 = 3.3565 ph= 14 - poH = 14-(pko + log (6.000) = 10.995 pos= pkb 7 og L Base