Question

In: Chemistry

32. Calculate the [H+]of the ammonium hydroxide solution from a pH of 10.63. mol/L= ? 33....

32. Calculate the [H+]of the ammonium hydroxide solution from a pH of 10.63.

mol/L= ?

33. Display the ICE table used to calculate your experimental Kb for ammonium hydroxide based on your values.

Solutions

Expert Solution

32)

pH = 10.63

[H+] = 10^-pH

         = 10^-10.63

          = 2.34 x 10^-11 M

[H+] = 2.34 x 10^-11 M

33)

[H+] = 2.34 x 10^-11 M

[OH-] = Kw / [H+]

         = 1.0 x 10^-14 / 2.34 x 10^-11

        = 4.27 x 10^-4 M

[OH-]= 4.27 x 10^-4 M = x

NH4OH (aq) ----------------> NH4+(aq) + OH-(aq)

C                                           0                  0 ---------> initial (I)

-x                                        +x                +x ----------> change (C)

C-x                                      x                  x ------------> equilibrium (E)

Kb = [NH4+][OH-]/[NH4OH]

Kb = x^2 / C-x

here x is know from above problem but initial concentration of NH4OH is not given so it is needed to calculate Kb

    


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