Question

complete the following chart with the appropriate responses Solution pH [H301 (OH) рон Acidic, Basic, or Neutral Strong acid 0.15 M HCI -log(0.15) 10.824 0.15 M 14.0-0.824 13,176 Strong auid 2.5 x 10 - MHCI 1-log(2.5.10-3) 4.100 2.5*10 14.0-4.60 9,4 weak acid 0.25 M HF dH2O weak base 10.0 M NH3 Strong base 2.7 x 10 - M KOH 14.0-3.57 10,43 2,7*10-4 /-log(2.7410-4) 3.57 Strong base 14,0-1824 0.15 M NaOH 13,186 0,1sm l-log (0,15) .824

Answer 1

Formula used: PH _-log [H30+) POH = -log [or] [PH + POH = -14 [Mt [OH-] = 1074

0.15M HR [ht] = [ce] = 0.15M. • pH = -log [ht] = -log 0:15 = 0.8 . (H20*] = [ht] = 0.15M [04] [ht] = kw. (oh) = kung = 1x1014 - 6.67 x 1014 M KW 4 S. 67 x 10 (H+) 0.15 epoh - -log[om-] = -log (6.67x10-14) = 13.18 TOR POH = 14 - PH = 14-0.82 = 13:18 opH <7 Acidic. # 2.5x105 M HCl. TH+] = 2.5x10-5. pH = -log (2.5 x10-5) = 4.6 o [H30+] = [ht] = 2.5x10-5 M. • For] = 10-14 - 10-14 = 4.0x10 10m 대버 2.5x10-5 POH = 14-pH = 14-4.6 = 9.4 • PH <7 Audic.

H₂O HF + I 0.25M cl-x E 0.25-X. H₂ot + f. 0 C tx + x x Ka = [F] [Hot [HE ka = x² 0.25-X x2 0.25 ka for HF = 6.6*10-4 (istandard value) x = Jo. 6x104x0.25 = 1.28 x 102m. [H₃O+] = a = 1.28 x 102 M. pH = -log [H30t] = -log(1-28 x1032) [PH = 1089] [H3O+] = 1.28 x 102M. ToH ] = 10-14 1028.x10-2. 7.81 x 10 e Pol - 14 - 1.89 = 12.11. • PH < 7 Acidic.

# d H2O : [H+][on] - 10-14. [ht] = [OH = 10-14 = 107. [Hot] =107 M. basses PH = -log[hzot] = -10g (1077) = 7. • Toh = 107 e pOH = -log (10-7) = 2 - PH = 1 (Neutral). 10.0M NH₃ NH₃ + H₂O CoM kb Nhut o + oH Kb = 1.8x105. (standard) MARECER -X ta +2 . 1o-x x x Tko = [Nha+] LOH [NH₃] ] = x² 1o-x = x² lo. x = Skb. 10 = 21.8x105 x 10 = 1.34x10²m. LOH] = 1.34 x 102 M. • 4₂0t1 = - - 7.46 x 101m 1019 [OH- 10-14 1634x10² pH = -log (143047 -_-log (7.46 x 103 - 12.13. (Basic) o. POH = 14-12.13 = 1.87.