At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10-11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?

Question

At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10-11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ? (A) 9 (B) 10 (C) 11 (D) 8

Tardigrade Admin · Accepted Answer

Mg 2+ + 2OH- leftharpoons Mg(OH)2 Ksp = [Mg2+][OH-]2 [OH-] = √(Ksp/[Mg2+]) = 10-4 ∴ pOH = 4 and pH = 10

Q. At $2 5^{\circ} C$ , the solubility product of $M g (O H)_{2}$ is $1.0 \times 1 0^{- 11}$ . At which pH, will $M g^{2 +}$ ions start precipitating in the form of $M g (O H)_{2}$ from a solution of $0.001 M M g^{2 +}$ ions ?

9

10

11

8

$M g^{2 +} + 2 O H^{-} ⇌ M g (O H)_{2}$
$K_{s p} = [M g^{2 +}] [O H^{-}]^{2}$
$[O H^{-}] = \frac{K _{s p}}{[ M g ^{2 +} ]} = 1 0^{- 4}$
$∴ p^{O H} = 4$ and $p^{H} = 10$

Q. At 25∘C, the solubility product of Mg(OH)2​ is 1.0×10−11. At which pH, will Mg2+ ions start precipitating in the form of Mg(OH)2​ from a solution of 0.001MMg2+ ions ?

9

10

11

8

Mg2++2OH−⇌Mg(OH)2​ Ksp​=[Mg2+][OH−]2 [OH−]=[Mg2+]Ksp​​​=10−4 ∴pOH=4 and pH=10

Q. At $2 5^{\circ} C$ , the solubility product of $M g (O H)_{2}$ is $1.0 \times 1 0^{- 11}$ . At which pH, will $M g^{2 +}$ ions start precipitating in the form of $M g (O H)_{2}$ from a solution of $0.001 M M g^{2 +}$ ions ?

$M g^{2 +} + 2 O H^{-} ⇌ M g (O H)_{2}$
$K_{s p} = [M g^{2 +}] [O H^{-}]^{2}$
$[O H^{-}] = \frac{K _{s p}}{[ M g ^{2 +} ]} = 1 0^{- 4}$
$∴ p^{O H} = 4$ and $p^{H} = 10$