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Equal volumes of 0.14 mol L–1 NH3 (Kb = 1.8 x 10–5) and 0.14 mol L–1...

Equal volumes of 0.14 mol L–1 NH3 (Kb = 1.8 x 10–5) and 0.14 mol L–1 HCl are mixed. Calculate the pH of the resulting solution.

(A) 2.9

(B) 5.0

(C) 5.2

(D) 6.2

(E) 7.8

Expert Solution

NH3 + HCl ------------> NH4Cl

no of moles of NH3 = 0.14moles

no of moles of Hcl = 0.14 moles

equal volume of NH3 and HCl

total volume = V+ V = 2v = 2

molairy of NH4^+ = 0.14/2 = 0.07 M

NH4^+ + H2O -------------. NH3 + H3O^+

I 0.07 0 0

C -x +x +x

E 0.07-x +x +x

Ka = Kw/Kb

= 1*10^-14/1.8*10^-5 = 5.6*10^-10

Ka = [NH3][H3O^+]/[NH4^+]

5.6*10^-10 = x*x/(0.07-x)

5.6*10^-10*(0.07-x) = x^2

x = 6.26*10^-6

[H3O^+] = x = 6.26*10^-6M

PH = -log[H3O^+]

= -log6.26*10^-6

= 5.2034 >>>>answer

queen_honey_blossom answered 2 years ago

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Question

Equal volumes of 0.14 mol L–1 NH3 (Kb = 1.8 x 10–5) and 0.14 mol L–1...

Solutions

Expert Solution

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