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What is the expected pH of 0.50 M NH3 (Kb = 1.8 x 10-5)?

What is the expected pH of 0.50 M NH3 (Kb = 1.8 x 10-5)?

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Homework Answers

Answer #1

Let us consider the reaction

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

Now Kb = ([NH4+)][OH-]/[NH3] (considering [H2O] =1)

Now let us consider X mole of NH4+/OH- is formed during hydrolysis. Therefore, (0.5-X) mole of NH3 remains.

[NH4+]=[OH-]=X anf [NH3]=0.5-X

Therefore, Kb= X.X/(0.5-X)

or Kb = X2 /0.5 (X is small therefore 0.5-X is nearly 0.5)

or X = /2

X = 0.00095

Now pOH= -log[OH-] = -log[0.00095] = 3.02

therefore pH= 14-pOH = 14-3.02 = 10.98

All the best and Cheers!!

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