Question

V.2000 m What IMM 0000.01 17. A 3.75 M solution of concentrated KOH is used to prepare 250.0 mL o volume of concentrated KOH is needed? (pick closest) O is used to prepare 250.0 mL of 0.180 M KOH. About what a. 2.0 mL b. 10.0mL C. 30.0 mL d. 70.0 mL Cox 2.1 e. 100.0 mL 18. What is the pH of a buffer that is 0.50 M HCN and 0.150 M NaCN? (The ka for HCN = 6.2 x 10-19) a. 9.19 + 10 Hq ortal W HOX MOS.O to mo.2s of bobbs al DOH MOS.O lo slamsejm.os. b. 4.81 (ox2.=) c. 4.29 (OH +(ps) 05 (ps)HOX + (pa)DOH d. 9.71 200. 0 0 0.0 slom sitt 100.0 + 200.0- 2000 - 3160 19. The solubility of Mg(OH)2 is 3.72 x10 SM. The solubility product Ksp is. 000.0 a. 1.38 x109 - 6.2.06 AOS C. 5.14 x10-14 sritgnizu Hq enimotob.926d nouze 1 HOX d. 2.68 x10-10 MSSO.O = (JM0.0\100,0) = h e. 3.07 x x10-5 DESI = 20.1-6 Haza.IH

Answer 1

Here the KOH solution et désired strength can be prepared by uning V.S, = V₂ S2 equation, let, v= 250.0m S = 0. 180 (M) S₂2 3.75 (M) 50 Vo = VLX SL V required volume of to 32 ott of Strength 3.75(M) = 250.0% 0-180 3.75 - 12 mL (which is closest to 10mL) so the required Kot is 12mL 3.75 (M).

Here the calculated amount of KOH is 12 mL but there is no option of 12 mL.So we should go with the nearest option of 12 mL i.e. 10 mL.

18. Hore HON as an weak acid. The NaCN is the highly dissociated sett ef weak acid HEN. And pH of such a solution. is giren by the equation pH pka't log [salt] pH = -loy [A] Tacid pka = -log ka ka= 6.2x10-10) her N = 6.5 M = -log (6.2x10-10) [NacN] = 0.15 m = 9.20 . - PH = 2.20 + b ( 0.5) 10.5) i = 8.684 (Answer)

19) It the solubility of Mylow)2 =s then, Mg 104)2 = Mg2+ + 2401 Ksp = {'Mg2+] {2 [ 02 ] Kapsclubility product is IN Sx (2522 453 - 4x (3.72x10-513 r. Saf Malot 23.12x10-5(M) = 2.109x10-13 (Answer)