The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10-9) is

Question

The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10-9) is (A) 0.0060% (B) 0.013% (C) 0.77% (D) 1.6%

Tardigrade Admin · Accepted Answer

α=√( K b / C ) =√(1.7 × 10-9/0.1) =√1.7 × 10-8 =1.3 × 10-4 =1.3 × 10-2 %

Q. The percentage of pyridine $(C_{5} H_{5} N)$ that forms pyridinium ion $(C_{5} H_{5} N^{+} H)$ in a $0.10 M$ aqueous pyridine solution ( $K_{b}$ for $C_{5} H_{5} N = 1.7 \times 1 0^{- 9}$ ) is

0.0060%

0.013%

0.77%

1.6%

$α = \frac{K _{b}}{C}$

$= \frac{1.7 \times 1 0 ^{- 9}}{0.1}$

$= 1.7 \times 1 0^{- 8}$

$= 1.3 \times 1 0^{- 4}$

$= 1.3 \times 1 0^{- 2} %$

Q. The percentage of pyridine (C5​H5​N) that forms pyridinium ion (C5​H5​N+H) in a 0.10M aqueous pyridine solution (Kb​ for C5​H5​N=1.7×10−9) is