Question

The pH of an aqueous solution of 0.328 M trimethylamine (a weak base with the formula (CHN) is

Answer 1

Solution (CH3), Mads + H2O4= (CM3), Nutraq) + Onicom) Initial 0.328 Change - Equilibrium 0.328-32 Al 280 Kolcus), w) = 65x105 Kb = [(cm) witCom] [cmg)NJ 6.5x10-5 = xxzi 0.328-3 Assuming n << 0.328 since ((13) ₂N is a weak base. 6.5x10-5 a ? : 0-328 >> x = 4.62 x 10 Ñ = Com] pOH = -log Com] = - log (4.62 x10-3) s 2.34 PH + POH = 14 3) PH2 14 -2.34 | PH₂ 11.66] Answer

2. I NH₂OH(aq) + H2010) NH₂Oht (aq) + On Cool Initial . 0.493 ... Change Equilibrium - 0.493- At 25°c k (NH2OH) 2.1.07 810-8 1 Kb : ENHgon!] [on] S .. [NHOH] 10TXIOS = 2² ...:: :0.493-%. . De 0.493 X1.07x10-8: acc 0.493, becausé 7 L NH₂OH is a weak base] :X: 7.26 X10--. Come] POH = -log[on] - logi (7.26 X10-) = 4.14 .pH = 14-4.14 39.86 PH² - log [rizot - [н, о*1, 10 еп, о 1 [hot] = 1.38*10-10 r. 1. Answer - 9:8.6 -

3. ((13) NH Saq, + OH 70q) CHE), Nagy? H2Ol) = 1 Initial 0.445 . Change : - Equilibrium 0.445-2 1 6:445-26 25°c Kyz .6.5x10-S : Кь .. . ... . . XS 0:445x 6.5X100 5 7x220.445] -> X = 5.38 x103 M2 [or] POM= -log Com] = -log(s.39X10°3) pOH = 2.27] Answer