[Four-digit Integer] An aqueous solution of aniline of concentration 0.2 M is prepared. How many milligrams of should be added in 500 ml of this solution so that aniline ion concentration in the solut

Question

[Four-digit Integer] An aqueous solution of aniline of concentration 0.2 M is prepared. How many milligrams of $NaOH$ should be added in 500 ml of this solution so that aniline ion concentration in the solution becomes 10^-8 M? Kb of C6H5NH2 = 4.0 × 10^-10.

Prashant Kotian · Accepted Answer

Correct answer: 0160. C6H5NH2 + ⇌ C6H5+ + OH- (Kb = 4.0 × 10^-10).
Kb = [C6H5+][OH-] / [C6H5NH2] => 4.0 × 10^-10 = (10^-8 × [OH-]) / 0.2 => [OH-] = 8.0 × 10^-3 M.
Since is a strong base, [OH-] approx [] = 8.0 × 10^-3 M.
Moles of in 500 mL = 8.0 × 10^-3 × 0.5 = 4.0 × 10^-3 mol.
Mass in mg = 4.0 × 10^-3 mol × 40 g/mol × 1000 mg/g = 160 mg.

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