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Solubility Product

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A solution has $0.05 M$ $M g^{2 +}$ and $0.05 M$ $N H_{3}$ . Calculate the concentration of $N H_{4} C l$ required to prevent the formation of $M g (O H)_{2}$ in solution.
$K_{s p} M g (O H)_{2} = 9 \times 10^{- 12}$ and ionisation constant of $N H_{3}$ is $1.8 \times 10^{- 5}$

A

0.067 M

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B

0.67 M

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C

0.01 M

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D

0.09 M

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Solution

The correct option is B $0.067 M$
The minimum $[⊝ O H]$ at which there will be no precipitation of $M g (O H)_{2}$ can be obtained by
$K_{s p} = [M g^{2 +}] [⊝ O H]^{2}$
$9.0 \times 10^{- 12} = [0.05] [⊝ O H]^{2}$
$∴ [⊝ O H] = 1.34 \times 10^{- 5} M$
Thus, a solution having $[⊝ O H] = 1.34 \times 10^{- 5} M$ will not show precipitation of $M g (O H)_{2}$ in 0.05 M $M g^{2 +}$ solution. These hydroxyl ions are to be derived by a buffer of $N H_{4} C l$ and $N H_{4} O H$ , i.e.,
$N H_{4} O H ⇌ N H_{4}^{\oplus} + ⊝ O H$
$N H_{4} C l ⟶ N H_{4} + C l^{⊝}$
For $N H_{4} O H, K_{b} = \frac{[N H_{4}^{\oplus}] [⊝ O H]}{[N H_{4} O H]}$
In presence of $N H_{4} C l$ ; all the $[N H_{4}^{\oplus}]$ are provided by $N H_{4} C l$ since common ion effect decreases dissociation of $N H_{4} O H$ .
$∴ 1.8 \times 10^{- 5} = \frac{[N H_{4}^{\oplus}] [1.34 \times 10^{- 5}]}{[0.05]}$
$∴ [N H_{4} \oplus] = 0.067 M o r [N H_{4} C l] = 0.067 M$

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Similar questions

Q. A solution has

0.05 M M g^{2 +}

0.05 M N H_{3}

. Calculate the concentration of

N H_{4} C l

required to prevent the formation of

M g (O H)_{2}

in this solution.

K_{s p} of M g (O H)_{2} = 9.0 \times 10^{- 12}

and ionization constant of

N H_{3} = 1.8 \times 10^{- 5}

\sqrt{1.8} = 1.34

)

Q. To a solution of

0.1 M

{M g}^{2 +}

0.8 M

{N H}_{4} C l

, an addition of equal volume of

{N H}_{3}

gives precipitate.

[{N H}_{3}]

in solution is:

K_{{s p}_{M g {(O H)}_{2}}} = 1.4 \times 10^{- 11}

K_{b_{{N H}_{4} O H}} = 1.8 \times 10^{- 5}

Q. To a solution of

0.1 M

M g^{2 +}

0.8 M

N H_{4} C l

, an equal volume of

N H_{3}

is added which just gives precipitate. Calculate

[N H_{3}]

K_{s p}

M g (O H)_{2} = 1.4 \times 10^{- 11}

K_{b}

N H_{4} O H = 1.8 \times 10^{- 5}

Q. Find moles of

N H_{4} C l

required to prevent

M g (O H)_{2}

from precipitating in a litre of solution which contains 0.02 mole of

N H_{3}

and 0.001 mole of

M g^{2 +}

K_{b} (N H_{3}) = 10^{- 5}; K_{s p} [M g (O H)_{2}] = 10^{- 11}

Q. Find moles of

N H_{4} C l

required to prevent

M g (O H)_{2}

from precipitating in a litre of solution which contains 0.02 mole of

N H_{3}

and 0.001 mole of

M g^{2 +}

K_{b} (N H_{3}) = 10^{- 5}; K_{s p} [M g (O H)_{2}] = 10^{- 11}

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