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Gay Lussac's Law, Avagadro's Law

At 50C, liqui...

Question

At 50C, liquid ${N H}_{3}$ has ionic product is $10^{- 30}$ . How many amide $(N H_{2^{-}})$ ions are present per ${m m}^{3}$ in pure liquid ${N H}_{3}$ ? (take $N_{A} = 6 \times 10^{23}$ )

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Solution

Ionic product of $N H_{3}, K_{N H_{3}} = [N H_{4}^{+}] [N H_{2}^{-}] = 10^{- 30}$

$[N H_{2}^{-}] = \sqrt{10^{- 30}} = 10^{- 15} M$

Since,

1 litre contains $10^{- 15} m o l e s$

then $10^{6} m m^{3}$ contains $10^{- 15} m o l e s$

Therefore $1 m m^{3}$ contains $= \frac{10^{- 15}}{10^{6}} = 10^{- 21} m o l e s$

Number of $N H^{2 -} i o n s = 6 \times 10^{23} \times 10^{- 21} = 600 i o n s$

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Similar questions

- 50^{\circ} C

N H_{3}

10^{- 30}

(N H_{2}^{-})

m m^{3}

N H_{3}

N_{A} = 6 \times 10^{23})

50^{o} C,

N H_{3}

K_{N H_{3}} = [N H_{4}^{+}] [N H_{2}^{-}] = 10^{- 30} M^{2}

N H_{2}^{-}

m m^{3}

50^{o} C,

N H_{3}

K_{N H_{3}} = [N H_{4}^{+}] [N H_{2}^{-}] = 10^{- 30} M^{2}

N H_{2}^{-}

m m^{3}

N H_{3}

2 N H_{3} ⇌ N H_{4}^{\oplus} + N H_{2}^{⊝}

K_{N H_{3}} = [N H_{4}^{\oplus}] [N H_{2}^{⊝}] = 10^{- 30} a t - 50^{o} C

(N H_{2}^{⊝})

m m^{3}

N H_{3}

N H_{3}

K_{S I C (N H_{3})} = 10^{- 30}

N H_{4}^{+}

100 c m^{3}

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