A buffer is 0.10 M in NH4OH and 1.0 M in NH4Cl (K_b = 1.8x10^-5). A solution containing 0.010 M each of Mg^2+ and Al^3+ is made up in this buffer. Take K_sp: Mg(OH)2 5.6x10^-12, Al(OH)3 1.3x10^-33. Which statements are correct?

[OH-] = 1.8x10^-6 M in the buffer

If NH4Cl were absent (0.10 M NH4OH alone), Mg(OH)2 would precipitate

[OH-] = 1.8x10^-6 M in the buffer

If NH4Cl were absent (0.10 M NH4OH alone), Mg(OH)2 would precipitate

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3qc197

A buffer is $0.10 M$ in $NH X_{4} OH$ and $1.0 M$ in $NH X_{4} Cl$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ). A solution containing $0.010 M$ each of $Mg X^{2 +}$ and $Al X^{3 +}$ is made up in this buffer. Take $K_{s p}$ : $Mg (OH) X_{2}$ $5.6 \times 1 0^{- 12}$ , $Al (OH) X_{3}$ $1.3 \times 1 0^{- 33}$ . Which statements are correct?

Buffer:

[OH X^{-}] = K_{b} \times 0.10/1.0 = 1.8 \times 1 0^{- 6} M

. For Mg:

0.010 \times (1.8 \times 1 0^{- 6})^{2} = 3.2 \times 1 0^{- 14} < 5.6 \times 1 0^{- 12}

, no precipitate. For Al:

0.010 \times (1.8 \times 1 0^{- 6})^{3} = 5.8 \times 1 0^{- 20} > 1.3 \times 1 0^{- 33}

, precipitate. Without NH4Cl,

[OH X^{-}] = 1.8 \times 1 0^{- 5} \times 0.10 = 1.3 \times 1 0^{- 3}

and the product

1.8 \times 1 0^{- 8} > 5.6 \times 1 0^{- 12}

, so Mg(OH)2 precipitates. Options 1, 2 and 4.