At what pH will $C u (O H)_{2}$ start to precipitate from a solution with $[C u^{2 +}] = 0.0015 M$ ?
$(K_{s p}$ for $C u (O H)_{2} = 1.5 \times 10^{- 19})$

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

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

9.4

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

4.6

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

Open in App

Solution

The correct option is C 6
The expression for the solubility product $(K_{s p})$ for $C u (O H)_{2}$ is $(K_{s p}) = [C u^{2 +}] [O H^{-}]^{2}$
But $[C u^{2 +}] = 0.0015 M$ and $(K_{s p}) = 1.5 \times 10^{- 19}$
Substitute values in the above expression.
$1.5 \times 10^{- 19} = 0.0015 \times [O H^{-}]^{2}$
Hence, $[O H^{-}] = 1 \times 10^{- 8}$
$p O H = - l o g [O H^{-}] = - l o g (1 \times 10^{- 8}) = 8$
$p H = 14 - p O H = 14 - 8 = 6$

Suggest Corrections

Similar questions

Q. The standard reduction potential for

C u^{2 +} / C u

+ 0.34

V. What will be the reduction potential at pH

= 14

? [Given:

K_{s p}

C u (O H)_{2}

1.0 \times 10^{- 19}]

$K_{s p}$ of $C u (O H)_{2}$ is 5 $\times 10^{- 16}$ at $25^{\circ} C$ , then pH of a saturated solution of $C u (O H)_{2}$ is___

Q. For

C u (O H)_{2}, K_{s p} = 1.6 \times 10^{- 19}

, the molar solubility of

C u (O H)_{2}

is:

The standard reduction potential for $C u^{2 +} | C u$ is $+ 0.34 V$ . Calculate the reduction potential at pH $= 14$ for the above couple if $K_{s p}$ of $C u (O H)_{2}$ is $1.0 \times 10^{- 19}$ .

Q. When caustic soda is added to an aqueous solution of copper sulphate, a blue precipitate of

C u (O H)_{2}

is obtained. How will you separate

C u (O H)_{2}

from the mixture ?

Join BYJU'S Learning Program

At what pH will Cu(OH)2 start to precipitate from a solution with [Cu2+]=0.0015M?(Ksp for Cu(OH)2=1.5×10−19)

At what pH will $C u (O H)_{2}$ start to precipitate from a solution with $[C u^{2 +}] = 0.0015 M$ ?
$(K_{s p}$ for $C u (O H)_{2} = 1.5 \times 10^{- 19})$