M^2+ (0.10 M) is to be removed from solution as M(OH)2 (K_sp = 2.5 x 10^-13) until only 0.1% of it remains. Using log 5 = 0.699, the minimum pH required is closest to

JEE Advanced levelSingle correct optionDifficulty 3 of 3ie201

$M X^{2 +}$ ( $0.10 M$ ) is to be removed from solution as $M (OH) X_{2}$ ( $K_{s p} = 2.5 \times 1 0^{- 13}$ ) until only $0.1%$ of it remains. Using $lo g 5 = 0.699$ , the minimum pH required is closest to

Final

[M X^{2 +}] = 0.1 \times 0.001 = 1.0 \times 1 0^{- 4} M

[OH X^{-}] = 2.5 \times 1 0^{- 13} /1 0^{- 4} = 5 \times 1 0^{- 5} M

, so

pOH = 5 - 0.699 = 4.301

and pH

= 9.70