Ionic EquilibriumHard
Question
Liquid ammonia ionizes to slight extent. At −50°C, its self-ionization constant, K = [NH4+] [NH2−] = 10–30 M2. How many amide ions are present per ml of pure liquid ammonia? (NA = 6 × 1023)
Options
A.10−15
B.10−18
C.6 × 105
D.6 × 108
Solution
$\left\lbrack NH_{2}^{-} \right\rbrack = \sqrt{10^{- 30}} = 10^{- 15}\text{ M}$
∴ Number of $NH_{2}^{-}$ ions per ml$= \frac{10^{- 15} \times 1}{1000} \times \left( 6 \times 10^{23} \right) = 6 \times 10^{5}$
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