Question

Equations pH=-log[H3O+]; pOH= -log[OH]; pKw=14.00=pH+POH; Ka-[H3O+][A-[HA]; Kb=[BH+][OH-)[B); pKa=-log Ka; Ka. Kb=Kw Constants Ka (HS-)=1x10-19; Ka (HF) 7.2x10-4; Ka ([Al(H20).]+)7.9x10-6; Ka (H3PO4)=7.5x10-2: Ka (HPO42-) =3.6x10-13; Ka (HCI)=Huge; Ka(Na+)=tiny; Ka(Cl-)tiny; Kb(NH3)=1.8x10-5; Kw=1x10-14 1) Calculate the pH of aqueous 0.25M HCl and 0.25M HF solutions. 2) Complete the following tables for aqueous solutions of conjugate acid-base pKa pKb Kb 1.3x10-4 35 3) Complete the following table for aqueous solutions DHL TH+) OH) pOH 1x102 4) Calculate the pH of aqueous 0.15M Ba(OH)2 and 0.15M NH; solutions.

Answer 1

1)
concentration of HCl = 0.25 M

pH = -log[H+]

   = -log0.25

   = 0.6

HF = weak acid

pH of weak acid = 1/2(pka-logC)

   pka of HF = 3.14

C = concentration of HF = 0.25 M

pH = 1/2(3.14-log0.25) = 1.87

2)

Ka           pKa           pKb                Kb

             (-logKa)     (pka+pkb=pkW)    (pkb = -logKb)

1.3*10^-4      3.9          10.1             7.94*10^-11

3.16*10^-11   10.5          3.5             3.16*10^-4

3) Kw = 1*10^-14 , pKw = 14

      pH            [H+]         [OH-]              pOH

(pH = -log[H+])             (Kw = [H+][OH-]   (pOH = -log[OH-])

      8            1*10^-8          1*10^-6         6

       12           1*10^-12        1*10^-2         2

4)

    Concentration of Ba(OH)2 , [OH-] = 0.15*2 = 0.3 M

    pOH = -log[OH-]

        = -log0.3

        = 0.523

   pH = 14-0.523 = 13.477

NH3 = weak base

pOH of weak base = 1/2(pkb-logC)

   pkb of NH3 = 4.74

C = concentration of NH3 = 0.15 M

pOH = 1/2(4.74-log0.15) = 2.78

pH = 14-2.78 = 11.22