What is the pH of a 0.0100 M sodium benzoate solution Kb for the benzoate anion (C7H5O2) = 1.5 1010 [Solved]

Let B^-1 = C7H5O2^-1
B^-1 H2O <> BH OH^-1
Kb = [BH][OH-] / [B^-1]
[BH] = [OH^-1]
1.510^-10 = [OH-]^2 / ([B^-1] [OH^-1])
Assume [OH-] << [B^-]
1.510^-10 = [OH-]^2 /[B-]
[OH-]^2 = 1.510^-10 x 0.0100
[OH-] = 1.2210^-6
pOH = 5.91
pH = 14.00 5.91 = 8.09

Let B^-1 = C7H5O2^-1
B^-1 H2O <> BH OH^-1
Kb = [BH][OH-] / [B^-1]
[BH] = [OH^-1]
1.510^-10 = [OH-]^2 / ([B^-1] [OH^-1])
Assume [OH-] << [B^-]
1.510^-10 = [OH-]^2 /[B-]
[OH-]^2 = 1.510^-10 x 0.0100
[OH-] = 1.2210^-6
pOH = 5.91
pH = 14.00 5.91 = 8.09

This post is last updated on hrtanswers.com at Date : 1st of September – 2022

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