Question

pelase help

5. (8 points) Calculate the pH of each of the following aqueous solutions at 25 C 0.65 M boric acid (B(OH), K,-7.3 x 10") a. 3.15 M ammonia (NH, K.- 1.76 x 10) b. c. 0.82 M benzoic acid (CaH,COOH, K.-6.3 x 10) 0.100 M HASO4 (K,-2.5 x 10, Ke 5.6 x 10, K-3.0 x 10) d. (2 points) Determine whether each of the following salts are acidic, basic, or neutral. 6. a. Na SO b. CH NH OI(K of CH,NH; = 4.4 x 10; K, of HOI - 2.3 x 10) с. КОВГ c. KOBr d CHNΗΝΟ,

Answer 1

a BCOH)3 + H20 H3O+ + BOCOH)2. kq- 7-3x10 ka= [H3O+][BO(OH)2] [BLOH)3 H₂O = BCOH)3 + 0.65M. H₃O+ + BOLOH)2 HOW + +20 0.65-x. 7.3x10 = C. 20 0.65-2 1-3x10-10 = 2 0.65 - Consider, 0.65-2=0.65 T.3x1010 = x2 0.65 22 - 7.3x100 x0.65 x = 4.745x10-10 x = 2.178 x 10-5 [+₂0+] = c = 2-178x105 pH = -log[H₂ot] = -10g 2.178X105 pH = 4.662 PH = 4.662

b. 815 M ammonia (NH, - 1-76*105) NH₃ + Hoo - NHAT + oH нош 3 .15 -3 3.15-x Kb = (NH4] [OH] (NH₃] kb = x. x 3.15-x consider 3.15-2 & 3.15. 1.T6x105 = x² 3-15 22 = 3.15*1.76x105 x = 85.544x10s x = 7.172 x 10-3 x= [OH] = 7.172x10 3 POH= -log[OH-] = -log (T.172x103). poh. = 2.144 .pH = 14 - pot = 14 -2.144 PH = 11.856 IpH = 11.856

C. 0.82M benzoic acid CHCOOH Kas 6.3x TOS C6Hgcoo +H30+. C6H5COOH + H2O - 0.62 How + 2 20 0.82-> ka= [C6H5 Coo] [H₃O+] [CEH5COOH) Κατ α . α (0.82-x) 6.3x105 = x2 (0.82-x) condider 0.82- 2 0-82. 6.3x105 = x2 0.82. x2 = 6.3x105x0.82 * = 5.166x10-5 x = 10.5166 x 104 x = 0.718T 10 2 = 7.187x10s x = [H₃O+] = 7-1878 103 pH = -log[H3O+] pH = -log (7.187*103) PH= 2.143 PH=2.143