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Solubility Product

At -50∘C, l...

Question

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

500 ions

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602.2 ions

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22 ions

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222.2 ions

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Solution

The correct option is B 602.2 ions
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, $1 L$ contains $10^{- 15}$ moles
$∴ 1 m m^{3}$ will contain $10^{- 15} / 10^{6}$ moles $= 10^{- 21}$ moles
number of $N H^{2 -}$ ins $= 6.022 \times 10^{23} \times 10^{- 21} = 602.0$ ions.

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Similar questions

Q. 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}

)

Q. At

50^{o} C,

the self-ionisation constant (ionic product) of

N H_{3}

is given as,

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

.
How many

N H_{2}^{-}

ions are present per

m m^{3}

of pure liquid ammonia?

Q. At

50^{o} C,

the self-ionisation constant (ionic product) of

N H_{3}

is given as,

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

.
How many

N H_{2}^{-}

ions are present per

m m^{3}

of pure liquid ammonia?

Q. Auto-ionisation of liquid

N H_{3}

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

with

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

Number of atomic ions

(N H_{2}^{⊝})

, present per

m m^{3}

of pure liquied

N H_{3}

is:

Q. Liquid

N H_{3}

ionises to a slight extent. At a certain temp. its self ionization constant

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

The number of

N H_{4}^{+}

ions are present per

100 c m^{3}

of pure liquid are:

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At −50∘C, liquid NH3 has ionic product is 10−30. How many amide (NH−2) ions are present per mm3 is pure liquid NH3? (Take NA=6×1023).

The correct option is B 602.2 ionsionic product of NH3=KNH3=[NH+4][NH−2]=10−30[NH−2]=√10−30=10−15MSince, 1L contains 10−15 moles∴1mm3 will contain 10−15/106 moles=10−21 molesnumber of NH2− ins=6.022×1023×10−21=602.0 ions.

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