Nicotine, $C_{10} H_{14} N_{2}$ , has two basic nitrogen atoms and both can react with water to give a basic solution
$N i c (a q) + H_{2} O (l) ⇌ N i c H^{+} (a q) + O H^{-} (a q)$
$N i c H^{+} (a q) + H_{2} O (l) ⇌ N i c H_{2}^{2 +} (a q) + O H^{-} (a q)$
$K_{b_{1}}$ is $8 \times 10^{- 7}$ and $K_{b_{2}}$ is $10^{- 10}$ . Calculate the approximate $p H$ of a $0.20$ M solution.

10.6

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11.65

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9.27

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12.11

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Solution

The correct option is B 10.6
$K_{b_{1}} = 8 \times 10^{- 7} = \frac{[N i c H^{+}] [O H^{-}]}{[N i c]}$
$[N i c]$ ..... $[N i c H^{+}]$ ..... $[O H^{-}]$
$I$ $0.2$ $0$ $0$
$- x$ $+ x$ $+ x$
$0.2 - x$ $x$ $x$
$∴ 8 \times 10^{- 7} = \frac{x^{2}}{0.2 - x} \Rightarrow x^{2} + 8 \times 10^{- 7} x - 1.6 \times 10^{- 7} = 0$
$∴ x = [N i c H^{+}] [O H^{-}] = 0.00039 = 3.9 \times 10^{- 4} M$

$k_{b_{2}} = 1 \times 10^{- 10} = \frac{[N i c H_{2}^{2 +}] [O H^{-}]}{[N i c H^{+}]}$
$[N i c H^{+}]$ ..... $[N i c H_{2}^{2 +}]$ ..... $[O H^{-}]$
$0.00039$ $0$ $0.00039$
$- x$ $+ x$ $+ x$
$0.00039 - x$ $x$ $0.00039 + x$
$∴ 10^{- 10} = \frac{x^{2} 1}{0.00039}$ [ $x$ in denominator can be ignored]
$\Rightarrow x^{2} = 3.9 \times 10^{- 4} \times 10^{- 10}$
$\Rightarrow x = 1.97 \times 10^{- 7}$
$∴ [O H^{-}] = 0.00039 + 1.97 \times 10^{- 7} = 0.00039 M$
$∴ p H = 14 - p O H$
$= 14 - 3.4 = 10.6$
$∴$ Approximate $p H = 10.6$

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Nicotine, C10H14N2, has two basic nitrogen atoms and both can react with water to give a basic solutionNic(aq)+H2O(l)⇌NicH+(aq)+OH−(aq)NicH+(aq)+H2O(l)⇌NicH2+2(aq)+OH−(aq)Kb1 is 8×10−7 and Kb2 is 10−10. Calculate the approximate pH of a 0.20M solution.