%3Cp%20class%3D%22MsoNormal%20c3%22%3E%3Cspan%20class%3D%22c1%20c2%22%3EAssume%2

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Explanation / Answer

NH3 + H2O <-----> NH4+ + OH-

Kb = 1.8 x 10^-5

1.8 x 10^-5 = x^2 / 0.11 -x

x = [OH-] =0.0014 M

pOH = - log 0.0014 = 2.9

pH = 14 - 2.9 =11.1

Moles NH3 = 0.0200 L x 0.11 = 0.0022

Moles HCl needed = 0.0022

V (HCl) = 0.0022 / 0.10 = 0.022 L

Total volume = 0.042 L

NH3 + H+ >> NH4+

NH4+ + H2O <----> NH3 + H3O+

K = Kw/ Kb = 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10

[NH4+] = 0.0022 / 0.042 = 0.052 M

5.6 x 10^-10 = x^2 / 0.052-x

x = [H3O+] = 5.4 x 10^-6 M

pH = 5.3

At the midpoint moles HCl needed = 0.0022 /2 = 0.0011

V ( HCl) = 0.0011 / 0.10 = 0.011 L

Total volume = 0.031 L

NH3 + H+ >> NH4+

Moles NH3 = 0.0022 - 0.0011 = 0.0011

Moles NH4+ = 0.0011

[NH3] = [NH4+] = 0.0011 / 0.031 =0.035

pOH = pKb + log [NH4+] / [NH3]

pKb = - log 1.8 x 10^-5 = 4.7

pOH = 4.7 + log 0.035 / 0.035 = 4.7

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