Question

What is the pH of a .62 M [C₅H₅NH]NO₃ solution?

K_b(C₅H₅N) = 1.7 x 10^-9

K_b(NH₃) = 1.8 x 10^-5

Answer 1

Dear Student

Given: Molar of pyridinium nitrate [CG H g NH] NOz solution = 0.62m Ko (CS HsN) = 1,78109 Ko CNHz) = 1.8x100s Now [CS H s NH] NO solution dissociates completely in mater to form pyridinium ion (Cg Hs. NH]* and nitrate ion No. [CS H & NH] NO ~ (CSHG NH]* + NO3 0.62m 0 0 0 0.62m 0.62m molarity of [Cg H , NH]* ion = 0.62m Nom, pyridinium ion does not dissociate completely [CS Hg NH]* Ho CS HSN + H₂O + Initial Change -X Equilibrium 0.62 - Here x is the change in concentration of [CS Hg NH]tion. Now, Ku - К. : К, =) Kas kw 1 Kb CO + H₂O 0.62 - 10-14 Now, - 1.78 109 - 5.882 x 10-6 ka = [CS HsN] [1₂0] [CH, NH*3 5.882 x 106 xxx 0.62 - [: 0x<< 0.62] 5.882 x 100 = x2 x x x 0.62

i 3) XP 0.62 x 5.882 8106 = 3.6468x10-6 2 = 1-9097 X10-3 [H30+] = 1,9097x10-3 pH = -log [Hot] E-log (1.4092x10-3) pH = 2.719 pH = 2.72 -
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