What will be the pH at which precipitation of Zn(OH)2 will take place from a solution containing 0.3 M Zn2+ ion? Ksp of Zn(OH)2 = 1.2 × 10-12

Question

What will be the pH at which precipitation of Zn(OH)2 will take place from a solution containing 0.3 M Zn2+ ion? Ksp of Zn(OH)2 = 1.2 × 10-12  (A)  5.7  (B)  8.3  (C)  11.4  (D)  6.2

Tardigrade Admin · Accepted Answer

Ksp of Zn(OH)2 =[Zn2+] [OH-]2 1.2×10-12=0.3×[OH-]2 [OH-]=2×10-6 pOH=-log [2×10-6]=5.699 ∴ pH=14-5.699 =8.301

Q. What will be the $p H$ at which precipitation of $Z n (O H)_{2}$ will take place from a solution containing $0.3 M Z n^{2 +}$ ion? $K_{s p}$ of $Z n (O H)_{2} = 1.2 \times 1 0^{- 12}$

$5.7$

$8.3$

$11.4$

$6.2$

$K_{s p}$ of $Z n (O H)_{2}$
$= [Z n^{2 +}] [O H^{-}]^{2}$
$1.2 \times 1 0^{- 12} = 0.3 \times [O H^{-}]^{2}$
$[O H^{-}] = 2 \times 1 0^{- 6}$
$pO H = - l o g [2 \times 1 0^{- 6}] = 5.699$
$∴ p H = 14 - 5.699$
$= 8.301$

Q. What will be the pH at which precipitation of Zn(OH)2​ will take place from a solution containing 0.3MZn2+ ion? Ksp​ of Zn(OH)2​=1.2×10−12

5.7

8.3

11.4

6.2