You visited us 500 times! Enjoying our articles? Unlock Full Access!

Heat of Formation

The pH at whi...

Question

The $p H$ at which $N i (O H)_{2}$ begins to precipitate from a solution containing $10^{- 3} M$ $N i^{2 +}$ ions:
$[K_{s p}$ of $(N i (O H)_{2} = 10^{- 11}]$

p H = 3

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

p H = 10

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

p H = 13

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

p H = 11

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

Open in App

Solution

The correct option is B $p H = 10$
$N i (O H) ⇌ N i^{2 +} + 2 O H^{-}$
$10^{- 3} \times [O H^{-}]^{2} = 10^{- 11}$
$[O H^{-}]^{2} = 10^{- 8}$
$[O H^{-}] = 10^{- 4}$
$p O H = 4$
$p H = 10$

Suggest Corrections

Similar questions

M g (O H)_{2}

M g^{2 +}

K_{s p}

M g (O H)_{2} = 1 \times 10^{- 11}

p H

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

0.10 M

{M g}^{+}

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

p H

M (O H)_{2}

0.10 M M^{2 +}

[K_{s p}

M (O H)_{2} = 1 \times 10^{- 9} 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}

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}

Join BYJU'S Learning Program