K sp of Mg OH 2-5-4 - 10-12 andK b for NH 3-1-8 - 10-5- K C for the reaction Mg OH 2 s -2 NH 4- 2 NH 3- H 2 O -Mg 2- isA- 3-33 - 10-7- 6-67 - 10-3C- 3-33 - 107D- 1-66 - 10-2

The correct option is D 1.66 × 10^ 2K_sp = [Mg^2+][OH^ ]^2/[Mg(OH)_2], K_b = [NH^+_4][OH^ ]; K = K_sp/K_b^2 = 5.4 × 10^ 12/(1.8 × 10^ 5)^2 = 1.66 × 10^ 2 ...

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Byju's Answer

Standard XII

Chemistry

Effective Equilibrium Constant

K sp of Mg OH...

Question

$K_{s p} o f M g (O H)_{2} = 5.4 \times 10^{- 12} a n d K_{b} f o r N H_{3} = 1.8 \times 10^{- 5} . K_{C}$ for the reaction $M g (O H)_{2} (s) + 2 N H_{4}^{+} \Leftrightarrow 2 N H_{3} + H_{2} O + M g^{2 +}$ is

3.33 \times 10^{- 7}

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6.67 \times 10^{- 3}

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3.33 \times 10^{7}

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1.66 \times 10^{- 2}

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Solution

The correct option is D $1.66 \times 10^{- 2}$
$K_{s p} = \frac{[M g^{2 +}] [O H^{-}]^{2}}{[M g (O H)_{2}]}, K_{b} = [N H_{4}^{+}] [O H^{-}]; K = \frac{K_{s p}}{K_{b}^{2}} = \frac{5.4 \times 10^{- 12}}{(1.8 \times 10^{- 5})^{2}} = 1.66 \times 10^{- 2}$

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Similar questions

Q. What is the

K_{c}

for the reaction:

{M g {(O H)}_{2}}_{(s)} + 2 {N H}_{4}^{+}_{(a q)} ⇌ 2 {N H}_{3}_{(g)} + 2 {H_{2} O}_{(a q)} + {M g}_{(a q)}^{2 +}

(Given that the solubility product for

M g (O H)_{2}, K_{s p}

6 \times 10^{- 12} m o l^{3} L^{- 3}

and

the base dissociation constant for

N H_{4} O H, K_{b}

1.8 \times 10^{- 5} m o l L^{- 1}

)

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 +}

ions.
Given :

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

Q. 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 this solution.

K_{s p}

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

and ionisation constant of

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

Q. 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 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}

(Given,

\sqrt{1.8} = 1.34

)

The solubility of

M g {(O H)}_{2}

is increased by the addition of

{N H}_{4}^{+}

ion.

Given:

K_{s p}

M g (O H)_{2} = 6 \times 10^{- 12}

K_{b}

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

Solubility of

M g {(O H)}_{2}

in a solution containing

0.5 M

{N H}_{4} C l

before addition of

M g {(O H)}_{2}

is:

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Ksp of Mg(OH)2=5.4×10−12andKb for NH3=1.8×10−5.KC for the reaction Mg(OH)2(s)+2NH+4⇔2NH3+H2O+Mg2+ is

The correct option is D 1.66×10−2Ksp=[Mg2+][OH−]2[Mg(OH)2],Kb=[NH+4][OH−];K=KspK2b=5.4×10−12(1.8×10−5)2=1.66×10−2

$K_{s p} o f M g (O H)_{2} = 5.4 \times 10^{- 12} a n d K_{b} f o r N H_{3} = 1.8 \times 10^{- 5} . K_{C}$ for the reaction $M g (O H)_{2} (s) + 2 N H_{4}^{+} \Leftrightarrow 2 N H_{3} + H_{2} O + M g^{2 +}$ is

The correct option is D $1.66 \times 10^{- 2}$
$K_{s p} = \frac{[M g^{2 +}] [O H^{-}]^{2}}{[M g (O H)_{2}]}, K_{b} = [N H_{4}^{+}] [O H^{-}]; K = \frac{K_{s p}}{K_{b}^{2}} = \frac{5.4 \times 10^{- 12}}{(1.8 \times 10^{- 5})^{2}} = 1.66 \times 10^{- 2}$