The Ksp of Mg(OH)2 is 1× 10-12. 0.01 M Mg(OH)2 will precipitate at the limited pH of

Question

The Ksp of Mg(OH)2 is 1× 10-12. 0.01 M Mg(OH)2 will precipitate at the limited pH of (A) 3 (B) 9 (C) 5 (D) 8

Tardigrade Admin · Accepted Answer

Mg ( OH )2 leftharpoons Mg 2++2 OH - The solubility product, K sp text of Mg ( OH )2 =[ Mg ]2+[ OH -]2 1 × 10-12 =[0.01][ OH -]2 [ OH -]2 =(1 × 10-12/0.01)=1 × 10-10 [ OH -] =10-5 ∵ [ H +][ OH -]=10-14 [ H +][10-5]=10-14 [ H +]=(10-14/10-5)=10-9 pH = - log [ H +] =- log 10-9=9

Q. The Ksp​ of Mg(OH)2​ is 1×10−12.0.01MMg(OH)2​ will precipitate at the limited pH of

3

9

5

8

Mg(OH)2​⇌Mg2++2OH− The solubility product, Ksp​ of Mg(OH)2​=[Mg]2+[OH−]2 1×10−12=[0.01][OH−]2 [OH−]2=0.011×10−12​=1×10−10 [OH−]=10−5 ∵[H+][OH−]=10−14 [H+][10−5]=10−14 [H+]=10−510−14​=10−9 pH=−log[H+] =−log10−9=9

Q. The $K_{s p}$ of $M g (O H)_{2}$ is $1 \times 1 0^{- 12} . 0.01 M M g (O H)_{2}$ will precipitate at the limited $p H$ of