The K_sp of a sparingly soluble hydroxide M(OH)2 is 3.2 x 10^-11. Taking log 2 = 0.301, the pH of its saturated aqueous solution at 298 K is closest to

JEE Main levelSingle correct optionDifficulty 2 of 3ie179

The $K_{s p}$ of a sparingly soluble hydroxide $M (OH) X_{2}$ is $3.2 \times 1 0^{- 11}$ . Taking $lo g 2 = 0.301$ , the pH of its saturated aqueous solution at 298 K is closest to

s = (K_{s p} /4)^{1/3} = 2 \times 1 0^{- 4} M

, so

[OH X^{-}] = 2 s = 4 \times 1 0^{- 4} M

pOH = 4 - 2 lo g 2 = 3.398

, hence pH

= 14 - 3.398 = 10.60