The percentage of pyridine -C-5H-5N- that forms pyridinium ion -C-5H-5N- H- in a 0-10 M aqueous pyridine solution -K-b C-5 H-5 N - 1-7 times 10-9- - is0-77 - 1-6-0-0060-0-013-

Question

The percentage of pyridine -C-5H-5N- that forms pyridinium ion -C-5H-5N- H- in a 0-10 M aqueous pyridine solution -K-b  C-5 H-5 N - 1-7 times 10-9- - is0-77 - 1-6-0-0060-0-013-

Toppr Admin · Accepted Answer

C5H5N-H2O-x2192-C5H5NH-OH-x2212-1Kb-C-x3B1-2-x3B1-x221A-KbC-x3B1-x221A-1-7-xD7-10-x2212-90-1-x3B1-1-30-xD7-10-x2212-4percentage-xA0-of-x3B1-1-30-xD7-10-x2212-4-xD7-100-0-013-Hence option D is the correct answer

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 $C_{5} H_{5} N = 1.7 \times 10^{- 9}$ ) is
$0.77 %$
$1.6 %$
$0.0060 %$
$0.013 %$

$C_{5} H_{5} N + H_{2} O \to C_{5} H_{5} N H^{+} + O H^{- 1}$

$K_{b} = C α^{2}$

$α = \sqrt{\frac{K_{b}}{C}}$

$α = \sqrt{\frac{1.7 \times 10^{- 9}}{0.1}}$

$α = 1.30 \times 10^{- 4}$

$p e r c e n t a g e o f α = 1.30 \times 10^{- 4} \times 100 = 0.013 %$

Hence option D is the correct answer

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 ) is0.77%1.6%0.0060%0.013%

C5H5N+H2O→C5H5NH++OH−1Kb=Cα2α=√KbCα=√1.7×10−90.1α=1.30×10−4percentage ofα=1.30×10−4×100=0.013%Hence option D is the correct answer

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 $C_{5} H_{5} N = 1.7 \times 10^{- 9}$ ) is
$0.77 %$
$1.6 %$
$0.0060 %$
$0.013 %$