K_sp of Cr(OH)3 is 2.7 x 10^-31. A solution contains 1.0 x 10^-3 M Cr^3+. Taking log 3 = 0.477, calculate the pH at which Cr(OH)3 just starts to precipitate (two decimals).

JEE Advanced levelNumerical answerDifficulty 3 of 3ie213

$K_{s p}$ of $Cr (OH) X_{3}$ is $2.7 \times 1 0^{- 31}$ . A solution contains $1.0 \times 1 0^{- 3} M$ $Cr X^{3 +}$ . Taking $lo g 3 = 0.477$ , calculate the pH at which $Cr (OH) X_{3}$ just starts to precipitate (two decimals).

Numerical answer type, tolerance 0.02.

[OH X^{-}]^{3} = K_{s p} / [Cr X^{3 +}] = 2.7 \times 1 0^{- 28}

, so

[OH X^{-}] = 3 \times 1 0^{- 10} M

pOH = 10 - 0.477 = 9.190

and pH

= 4.81