You visited us 1 times! Enjoying our articles? Unlock Full Access!

Solubility Product

MgOH2 is prec...

Question

$M g (O H)_{2}$ is precipitated when NaOH is added to a solution of $M g^{2 +}$ . If the final concentration of $M g^{2 +}$ is $10^{- 10}$ M, the concentration of $O H^{-} (M)$ in the solution is [Solubility product for $M g (O H)_{2} = 5.6 \times 10^{- 12}$ ].

0.056

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.24

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.025

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $0.24$
$M g (O H)_{2} \to M g^{2 +} + 2 O H^{-}$
$10^{- 10}$ C
Now,

$K_{s p} = [M g^{2 +}] [O H^{-}]^{2} = 10^{- 10} \times [O H^{-}]^{2} = 5.6 \times 10^{- 12}$

$\Rightarrow [O H^{-}] = 0.24 M$

Suggest Corrections

Similar questions

M g (O H)_{2}

N a O H

M g^{2 +}

M g^{2 +}

10^{- 10}

O H^{-}

[Solubility product for M g (O H)_{2} = 5.6 \times 10^{- 12}]

M g (O H)_{2}

9.0 \times 10^{- 12}

0.010

M g^{2 +}

M g (O H)_{2}

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

1 \times 10^{- 11}

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

{M g}^{2 +}

25^{o} C

M g (O H)_{2}

1.0 \times 10^{- 11}

p H

M g^{2 +}

M g (O H)_{2}

M g^{2 +}

M g (O H)_{2}

9 \times 10^{- 12}

M g^{+ 2}

0.01

O H^{-}

Join BYJU'S Learning Program