Question

all questions please

WER. Write the word or phrase that best completes each statement or answer the question KT ANSWER, W, 1.2) In order for the reaction 1. the Ka of HA must be for the reaction HAHSO3" -A- H2SO to have an equilibrium constant K (greater, less than the Ka of HSOY. 12) 3) The pH of a 0.045 M KOH solution MULTIPLE CHOICE Choose the Choose the one alternative that best completes the statement or answers the question. no a Lewis acid-base reaction forming an adduct. Which picture below ents the curved arrow notation for the initial Lewis acid-Lewis base interaction in this monction what is the Lewis acid and the Lewis base correctly represents the curved arrow ne 14) A) Picture (2) is correct; BF3 is the Lewis acid and NH3 is the Lewis base. B) Picture (3) is correct; NH3 is the Lewis acid and BF3 is the Lewis base. C) Picture (4) is correct; BF3 is the Lewis acid and NH3 is the Lewis base. D) Picture (1) is correct; NH3 is the Lewis acid and BF3 is the Lewis base. 15) 15) What is the resulting pH when 0.005 moles of KOH is added to 0.100 L of a buffer solution that is 0.100 M in H2PO4- and 0.100 M HPO42- and the Ka2 =6.2 10-8? A) 5.21 B) 7.69 C) 6.73 D) 5.61 16) Calculate the pH of a 1.60 M CH3NH3Cl solution. Kb for methylamine, CH3NH2, is 3.7 10-4 A) 1.61 B) 12.39 C) 8.82 D) 5.18 16) 17) Aniline, (C6H5NH2, Kb = 4.3 x 10-10 at 25°C) is an industrially important amine used in the making of dyes. Determine the pH of an aniline solution made by dissolving 6.90 g of aniline in enough water to make 100 mL of solution A) 4.75 B) 9.50 17) C) 10.50 D) 9.25

Answer 1

(2) HA + HSO - A + H₂SO . ,Kecil to obtain reg den than 1 cue) 1. Ke = keq = (ka) un (ка),зо, So, (ka) HA munks he less than the ka of HS og 13. [kon] = 0.045M Corel= 0.045m pole = -logy. Colt Da -hoo(0.045) = 1035. pH = 14-pola 12.65 . if (Co-ordinate Bond) ly. ONTH (Bf - Nh Adduct) Electron deficients Because of presence of Electron mich because of presence of lene pair of electrons and this act empty orbital and thus act as lawis acid of elections and th as leurs Base AD, (A) ü comect 1 152 CON NEXT PAÇE)

0.05m Skor ] ROH = Vill limiting reactants TORT+ H₂ pouť 0.005 moles Oile - tport + H₂o, keq - Kag Ab 0:05m DIM 6/ 11 Neutralisatim von 0.15M d. = 6,2 810-8 10th = 6-2 X10 0,05m = 6.2 X10%. ouis- 0.15-k 0.05th J Hydrolysis by HPQ2 No.05 ^o.ir is unity as 6-2 X 106 = kee 2 I C H P04] [017] oog? tinities of hero pure liquid is considered to be unity, 6. 2X106 = دکمه Cora to z X106 COŇ I = 0.484x10-6 pon = log [of] = 6:31. pH-14 - pon (at 29869 1 pH = 14-6131=7.69 Part B is correct 160 CH Nily + H₂O Ab tot loom Z CIENĄ + H₂0% kena kw ( -10-14 Aftate 1.6-4 3.7 X 70-4 1.6

* * * и 2 , . , , , — Х. "°. сочгах І. • C4, o"] => o» гт х/ [r* -4, C+0°) - - - rə x ) = г.) Part D is correct (19) Снен, , K, ч.3xtor' Aniline Моля мл = (x12+ +x 2 + 2 93 /my. [c., нек. ? " - Х/pos - 6.5 х Veme) = 6. -421 93 x — . ( К,уну яе ко - Кмн. , о = <a H-Nң + ой к, tako +4 2. Ч.3x 26 % о: 142- pok » - le cәй )4.2 ~e 2 ek ? /4 - poh : 19 -Ч.?r * о. 2 ч.3 xt-də грн = 8-27 (ои г яг , x1.-г. 1 Part d ů the ansu