Calculate the pH and moralities of all the species in a 0.150 M ammonia solution. Ka...

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Chemistry

Calculate the pH and moralities of all the species in a 0.150 Mammonia solution. Ka = 5.55 x 10-10

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3.6 Ratings (270 Votes)

Given :

[NH3] = 0.150 M

Ka = 5.55 E-10

Solution:

Kb is calculated as

Kb = 1.0 E-14 /ka = 1.0 E-14/5.55 E-10 = 1.8 E-5

ICE

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â NH3 Â Â Â Â + Â Â Â Â Â Â Â Â Â Â H2OÂ Â ----- > NH4+Â Â + OH-

IÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0.150-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 Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â 0

CÂ 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 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 Â +xÂ Â Â Â Â Â Â Â Â Â +x

E Â Â Â Â Â Â Â Â Â Â Â Â (0.150-x)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 Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â x

Ka = 1.8E-5 = x^2 / 0.150Â Â Â Â Â Â Â Â Â Â since ka is very small we can neglect x in the denominator.

x = 1.644 E-3

[OH-] = 1.644E-3 M

pOH = - log (1.644 E-3 ) = 2.87

pH=14-2.87 = 11.2

pH = 11.2

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