Question

2. Calculate the concentrations and activities of H3O+ for the following solutions (assume that the ionic strength is determined by the acid/base). Calculate the pH for each solution two ways: 1) pH = -log[H3O+] and 2) pH = -log @,0+. (a) 0.010 M HNO3 (b) 0.035 M KOH (c) 0.030 M HCI (d) 3.0 M HCI

Answer 1

Solalion! - 0.olom HNO2 we know that [H] = [HNO] = o.olom and [at] = [Hot] = o-olom pH = -log[HOT] - -dog (0.010) pH = -log(101= -log(102) = 2 and we also known ahot = [Hot] = 0.olom Thus - Clot) = 0.0lom ja hot = o.olom LPH = 2 0.035 m KOH we know - COH-] = [kohJ = 0.035M POH = -log Com-3 = -log(0.035) = 1-45 we know that pH & POH = 14 pH = 14 - pOH = 14-145 [PH = 12:55] pH = - dog [thot] 12.55= -log[hot] : [Hot] = 1012-55 - hiss = 9-81881013 m [hot] = 2.818 81013m) and apot = [hot] = 2.818810lm abot = 2.818x10 13m)

0.030 M HCl CHCd] =[HI] = 0.030m - [H₂ot] [Hot] -0.030m, ahot = [hot] = 0.030m pH = -log CH2043 = -log(0-030] = 1.52 pH = -log[amot) = -log(0.030) = 1.52 -- [pH = 1:52] [Hot] = 0.030m, 124407 = 0.030 3.om Hal we know that concentration of Ht lon = concentsation of HCl [ht] = Chel] = 3 m and [H0+] = [ht] - 3m, ahot = [Hot] = 3m pH = -log[hot] = -log 3 = -0.4771 .pH = 0.4771 Chot) – 3 m and Q4O+ = 3
Thank you!