The solubility product of a rare earth metal hydroxide $M {(O H)}_{3}$ at room temperature is $4.32 \times 10^{- 14}$ . Its solubility is:

2 \times 10^{- 3} M

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

2.0 \times 10^{- 4} M

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

2 \times 10^{- 5} M

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

2.0 \times 10^{- 6} M

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

Open in App

Solution

The correct option is A $2 \times 10^{- 3} M$
$M (O H)_{3} ⇌ M^{3 +} + 3 O H^{-}$
$K_{s p}$ for $A B_{3}$ type molecules is $K_{s p} = 27 s^{4}$
$4.32 \times 10^{- 14} = 27 s^{4} s^{4} = \frac{4.32 \times 10^{- 14}}{27} = 16 \times 10^{- 12} s = 2 \times 10^{- 3} M$

Suggest Corrections

Similar questions

Q. Calculate solubility of

A g C l

0.1 M N a C l

25^{o} C

if its solubility product at same temperature is

2.0 \times 10^{- 10} m o l^{2} L^{- 2}

Q. Calculate solubility of

A g C l

0.1 M N a C l

25^{o} C

if its solubility product at same temperature is

2.0 \times 10^{- 10} m o l^{2} L^{- 2}

The solubility of $C a F_{2}$ is $2 \times 10^{- 4}$ . Its solubilily product $K_{S P}$ is

Q. The solubility product of a sparingly soluble salt AB at room temperature is

1.21 \times 10^{- 6}

, its molar solubility is:

The solubility product of $B a S O_{4}$ is $1.5 \times 10^{- 9}$ . The precipitation in a $0.01 M$ $B a^{2 +}$ ions solution will start on adding $H_{2} S O_{4}$ of concentration -

Join BYJU'S Learning Program

The solubility product of a rare earth metal hydroxide M(OH)3 at room temperature is 4.32×10−14. Its solubility is:

The correct option is A 2×10−3MM(OH)3⇌M3++3OH−Ksp for AB3 type molecules is Ksp=27s44.32×10−14=27s4s4=4.32×10−1427=16×10−12s=2×10−3M

The solubility product of a rare earth metal hydroxide $M {(O H)}_{3}$ at room temperature is $4.32 \times 10^{- 14}$ . Its solubility is:

The correct option is A $2 \times 10^{- 3} M$
$M (O H)_{3} ⇌ M^{3 +} + 3 O H^{-}$
$K_{s p}$ for $A B_{3}$ type molecules is $K_{s p} = 27 s^{4}$
$4.32 \times 10^{- 14} = 27 s^{4} s^{4} = \frac{4.32 \times 10^{- 14}}{27} = 16 \times 10^{- 12} s = 2 \times 10^{- 3} M$