A solution is 0.010 M in Mn^2+ and contains 0.10 M NH3. Take K_b(NH3) = 1.8x10^-5 and K_sp[Mn(OH)2] = 2.0x10^-13. Does Mn(OH)2 precipitate (i) with no NH4Cl added and (ii) when 1.00 M NH4Cl is also present?

JEE Advanced levelSingle correct optionDifficulty 3 of 3qc101

A solution is $0.010 M$ in $Mn X^{2 +}$ and contains $0.10 M$ $NH X_{3}$ . Take $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ and $K_{s p} [Mn (OH) X_{2}] = 2.0 \times 1 0^{- 13}$ . Does $Mn (OH) X_{2}$ precipitate (i) with no $NH X_{4} Cl$ added and (ii) when $1.00 M$ $NH X_{4} Cl$ is also present?

(i)

[OH X^{-}] = K_{b} c = 1.8 \times 1 0^{- 6} = 1.34 \times 1 0^{- 3}

; ionic product

= 0.010 \times (1.34 \times 1 0^{- 3})^{2} = 1.8 \times 1 0^{- 8} > 2.0 \times 1 0^{- 13}

, so it precipitates. (ii)

[OH X^{-}] = 1.8 \times 1 0^{- 5} (0.10) /1.00 = 1.8 \times 1 0^{- 6}

; ionic product

= 0.010 \times (1.8 \times 1 0^{- 6})^{2} = 3.2 \times 1 0^{- 14} < 2.0 \times 1 0^{- 13}

, so no precipitate. The common-ion effect of

NH X_{4} Cl

keeps Mn in solution until the sulphide step.