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

Solubility Product

What is the m...

Question

What is the maximum $p H$ of $0.10 M$ solution in $M g^{2 +}$ from which $M g (O H)_{2}$ will not precipitate. Given : $K_{s p} (M g (O H)_{2}) = 1.2 \times 10^{- 11}$ .

Open in App

Solution

$M g (O H)_{2} ⇌ M g^{2 +} + 2 (O H)^{-}$
$K_{s p} = [H g^{2 +}] [O H^{-}]^{2}$
$[O H^{-}] = \sqrt{\frac{k_{s p}}{[{M g}^{2 +}]}} = \sqrt{\frac{1.2 \times 10^{- 11}}{0.1}} = 1.1 \times 10^{- 5} M$
$p O H = - l o g [O H^{-}] = - l o g (1.1 \times 10^{- 5}) = 5 - l o g 1.1 = 4.959$
$p H = 14 p O H ⟹ p H = 9.04$
Maximum $p H$ at which no precipitation occurs.

Suggest Corrections

Similar questions

0.10 M

{M g}^{2 +}

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

K_{s p} M g {(O H)}_{2} = 1.2 \times 10^{- 11} M^{3}

0.10 M

M g^{2 +}

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

K_{s p}

M g (O H)_{2} = 1.2 \times 10^{- 11} M^{3}

p H

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

0.10 M

{M g}^{2 +}

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

K_{s p}

M g (O H)_{2}

1.2 \times 10^{- 11}

M g^{2 +}

1.2 \times 10^{- 5} m o l / L

M g (O H)_{2}

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

1 \times 10^{- 11}

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

{M g}^{2 +}

Join BYJU'S Learning Program