The solubility product of N i( OH )2 at 298 K is 2 × 10-15 mol 3 ⋅ d m-9. The pH value if its aqueous and saturated solution is

Question

The solubility product of N i( OH )2 at 298 K is 2 × 10-15 mol 3 ⋅ d m-9. The pH value if its aqueous and saturated solution is (A) 5 (B) 7.5 (C) 9 (D) 13

Tardigrade Admin · Accepted Answer

N i(O H)2 leftharpoons underset S mol / LN i2++ underset2S mol/L2 OH - ⇒ Ks p=[N i2+][O H-]2=S ×(2 S)2=4 S3 =2 × 10-15( mol / L)3 =2 × 10-15 mol 3 d m-9 ⇒ S=((2/4) × 10-15)1 / 3=8 × 10-6 mol / L [O H-]=2 S=2 × 8 × 10-6 mol / L p O H=6- log 16=4.983=5 ∴ p H=14-p O H=14-5=9 at 25° C

Q. The solubility product of Ni(OH)2​ at 298K is 2×10−15mol3⋅dm−9. The pH value if its aqueous and saturated solution is

5

7.5

9

13

Ni(OH)2​⇌Smol/LNi2+​+2Smol/L2OH−​ ⇒Ksp​=[Ni2+][OH−]2=S×(2S)2=4S3 =2×10−15(mol/L)3 =2×10−15mol3dm−9 ⇒S=(42​×10−15)1/3=8×10−6mol/L [OH−]=2S=2×8×10−6mol/L pOH=6−log16=4.983=5 ∴pH=14−pOH=14−5=9 at 25∘C

Q. The solubility product of $N i (O H)_{2}$ at $298 K$ is $2 \times 1 0^{- 15} m o l^{3} \cdot d m^{- 9}$ . The $p H$ value if its aqueous and saturated solution is