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

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32)

pH = 10.63

[H+] = 10^-pH

Â Â Â Â Â Â Â Â = 10^-10.63

Â Â Â Â Â Â Â Â Â = 2.34 x 10^-11 M

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

33)

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

[OH-] = Kw / [H+]

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

Â Â Â Â Â Â Â = 4.27 x 10^-4 M

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

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

CÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0 ---------> initial (I)

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

C-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 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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