In: Chemistry
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.
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