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Find [NH3], [NH4+], [OH–], [H3O+] and pH of a 0.080 M NH3 solution NH3(aq) + H2O(l)...

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

Expert Solution

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

queen_honey_blossom answered 2 years ago

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