Question

a. What is the H+ concentration for an aqueous solution with pOH = 3.73 at 25 ?C?

b. Arrange the following aqueous solutions, all at 25 ?C, in order of decreasing acidity.

0.0023 M HCl, 0.0018 M KOH, pH = 5.45, pOH = 8.55

c. At a certain temperature, the pH of a neutral solution is 7.33. What is the value of Kw at that temperature?

Answer 1

a. POH   = 3.73

     PH    = 14-POH

               = 14-3.73

    PH     = 10.27

-log[H^+]   = 10.27

       [H^+]    = 10^-10.27   =5.4*10^-11M

b. POH   = 8.55

   PH    = 14-POH

            = 14-8.55

           = 5.45

     KOH(aq) ---------------> K^+ (aq) + OH^- (aq)

0.0018M                                            0.0018M

      [OH^-]    =    [KOH]

      [OH^-]    = 0.0018M

     POH    = -log[OH^-]

                 = -log0.0018

                   = 2.7447

    PH         = 14-POH

                  = 14-2.7447

                   = 11.2553

     HCl(aq) --------------> H^+ (aq) + Cl^- (aq)

0.0023M                        0.0023M

       [H^+]      =   [HCl]

      [H^+]      = 0.0023M

      PH     = -log[H^+]

                = -log0.0023

                 = 2.6382

0.0023MHCl>(POH =8.55) = (PH=5.45) > 0.0018 M KOH

c.

        PH   = 7.33

      -log[H^+]   = 7.33

           [H^+]     =10^-7.33   = 4.67*10^-8M

in neutral solution [H^+]   = [OH^-]   = 4.67*10^-8 M

   Kw   = [H^+][OH^-]

Kw     = 4.67*10^-8M * 4.67*10^-8M = 2.2*10^-15M^2