Question

Ch. 18 Values The following values will be useful for problems in this chapter. Acid Substance or Species HF HNO2 CH3COOH HOCI HOBr HOCN HCN H2SO4 KA = 7.2 x 10-4 Ka = 4.5 x 10-4 Ka = 1.8 x 10-5 Ka = 3.5 x 10-8 = 2.5 x 10-9 K4 = 3.5 x 10-4 Ka = 4.0 x 10-10 Kai = very large Kq2 = 1.2 x 10-2 Kal = 4.2 x 10-7 K2 = 4.8 x 10-11 NH3 CH3)2N [CO(OH2)6]2+ [Fe(OH2)]2+ [Fe(OH2)6]3+ [Be(OH)4]2+ [Cu(OH)4]2+ HBO2 (COOH)2 Kb = 1.8 x 10-5 K = 7.4 x 10-5 K = 5.0 x 10-10 KA = 3.0 x 10-10 Ka = 4.0 x 10-3 K4 = 1.0 x 10-5 K = 1.0 x 10-8 K = 6.0 x 10-10 Kal = 5.9 x 10-2 K22 = 6.4 x 10-5 K; = 5.0 x 10-4 H2CO3 CH3NH2

Explain the process of how to get the answer. The answer is highlighted.

Answer 1

Answer Cuday] =0.25m dissociation of Holau H2O + H clou - tato telou cu clou) 20.25 [cat] = 0.35)(cloy = 0.25 since (cusot] = 0.25 m.). The concentration (Hgot) will be equal to as caclou). from inic product of water (kw) = (hact] Cort] 2H20 a Hgot t out I kw a Cuzf] Cont], kw for water at 25° is given as 1.0x104 Kw=loxioly = Cugot] com] Cusut Cori] = loxiou 0.25 x Cori] = lo xcole . Con}= lo har 104 1. [Cour) = 4x10th a so option @ Chaut] =0.25m, cor]= urox10 m is correct Baco4), tras Bacousa is - ionized Bat2 + 2 into (or] one Bath and a each cou) ions (Ba Cou),] = 0.05m there are 2 out (Bata)= 0.05m ions so taatal cond= ax0 - USM (ou) = 0.1MY kw=Crizot] coni] = loxiola Cugut] (0.13 = 10 xiðlu 14+1 lo X10 eloxiduti 1- XOY lo X10 Tai - Cusot] = loxioly = 10x10 (Cuyot] = lo x103 mk so option ® is correct Answer