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If a buffer solution is 0.200 M in a weak base (Kb = 7.9

If a buffer solution is 0.200 M in a weak base (Kb = 7.9

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Answer #1

For buffer, we use the Henderson Hasselbach equation

pOH = pKb + log ([HB+]/[ B ])

The weab base is B and the protonated base is HB+

B+H2O <-> HB+ and OH-

Kb = [HB+][OH-] /[ B ]

Assume = [HB+]= [OH-] = x

[B] = 0.200 M -x (you may ignore this x since Kb is very small)

[B] = 0.2 M

Substitute in Kb

Kb = [HB+][OH-] /[ B ]

7.9*10^-5 = x*x / (0.2)

x^2 = (7.9*10^-5)(0.2) = 0.000158

x = sqrt(0.0000158) = 0.00397

[OH-] = x = 0.00397

pOH = -log(OH-) = -log(0.00397) = 2.4

but we need pH so

14 = pH+ pOH

pH = 14- 2.4 = 11.6

pH = 11.6

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