In: Chemistry
Find [NH3], [NH4+], [OH–], [H3O+] and pH of a 0.080 M NH3 solution
NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)
Kb = 1.76·10–5
NH3(aq) + H2O(l) ------------->NH4+(aq) + OH–(aq)
I 0.8 0 0
C -x +x +x
E 0.8-x +x +x
Kb = [NH4^+][OH^-]/[NH3]
1.76*10^-5 = x*x/0.8-x
1.76*10^-5*(0.8-x) = x^2
x = 0.00374
[NH4^+] = x= 0.00374M
[OH^-] = x = 0.00374M
[NH3] = 0.8-x = 0.8-0.00374 = 0.79626M
[H3O+] = Kw/[OH^-]
= 1*10^-14/0.79626 = 1.25*10^-14 M
PH = -log[H3O+]
= -log1.25*10^-14
= 13.9030 >>>>>answer