Question

The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅NH) in a 0.10 M aqueous pyridine solution (K_b for C₅H₅N = 1.7 × 10⁻⁹) is ____________.

Options

• 0.006%

• 0.013%

• 0.77%

• 1.6%

Answer 1

The percentage of pyridine (C₅H₅N) that forms pyridinium ion (C₅H₅NH) in a 0.10 M aqueous pyridine solution (K_b for C₅H₅N = 1.7 × 10⁻⁹) is 0.013%.

Explanation:

\[\ce{C5H5N + H - OH ⇌ C5H5\overset{+}{N}H + OH^-}\]

`(α^2"C")/(1 - α)` = K_b

α²C ≂ K_b

α = `sqrt("K"_"b")/"C" = sqrt(1.7 xx 10^-9)/0.1`

= `sqrt1.7 xx 10^-4`

Percentage of dissociation = `sqrt1.7 xx 10^-4 xx 100`

= 1.3 × 10⁻²

= 0.013%