250 mL of 0.16 M MgCl 2 ( aq ) is mixed with 750 mL of 0.2 M NH 3 ( aq ) resulting in precipitation of some Mg ( OH ) 2 . To redissolve Mg ( OH ) 2 , what minimum mass (in gram) of ( NH 4 )

Question

250 mL of 0.16 M MgCl 2 ​ ( aq ) is mixed with 750 mL of 0.2 M NH 3 ​ ( aq ) resulting in precipitation of some Mg ( OH ) 2 ​ . To redissolve Mg ( OH ) 2 ​ , what minimum mass (in gram) of ( NH 4 ​ ) 2 ​ SO 4 ​ should be added to the above solution? (Given : K SP [Mg(OH) 2 ] = 9 × 10 -12 , K b (NH 3 ) = 2 × 10 -5 )

TestHub · Accepted Answer

To redissolve Mg(OH)₂, the [OH⁻] must be lowered.

Initial moles of MgCl₂ = $0.16 M \times 0.250 L = 0.04 m o l$ .

Initial moles of NH₃ = $0.2 M \times 0.750 L = 0.15 m o l$ .

Total volume = $0.250 L + 0.750 L = 1.0 L$ .

So, $[M g^{2 +}] = 0.04 M$ and $[N H_{3}] = 0.15 M$ .

For Mg(OH)₂ to redissolve, $Q_{s p} < K_{s p}$ . At the point of redissolution, $Q_{s p} = K_{s p}$ .

$K_{s p} = [M g^{2 +}] [O H^{-}]^{2}$

$9 \times 1 0^{- 12} = (0.04) [O H^{-}]^{2}$

$[O H^{-}] = \frac{9 \times 1 0 ^{- 12}}{0.04} = 2.25 \times 1 0^{- 10} = 1.5 \times 1 0^{- 5} M$ .

The added (NH₄)₂SO₄ will react with NH₃ to form a buffer, reducing [OH⁻].

NH₃(aq) + H₂O(l) $⇌$ NH₄⁺(aq) + OH⁻(aq)

$K_{b} = \frac{[ N H _{4}^{+} ] [ O H ^{-} ]}{[ N H _{3} ]}$

Let $x$ be the moles of (NH₄)₂SO₄ added. This produces $2 x$ moles of NH₄⁺.

$2 \times 1 0^{- 5} = \frac{( 2 x ) ( 1.5 \times 1 0 ^{- 5} )}{0.15}$

$2 \times 1 0^{- 5} \times 0.15 = 3 x \times 1 0^{- 5}$

$0.3 \times 1 0^{- 5} = 3 x \times 1 0^{- 5}$

$x = 0.1 m o l$ .

Molar mass of (NH₄)₂SO₄ = $2 (14.01) + 8 (1.008) + 32.07 + 4 (16.00) = 132.14 g / m o l$ .

Mass of (NH₄)₂SO₄ = $0.1 m o l \times 132.14 g / m o l = 13.214 g$ .

Rounding to one decimal place, the minimum mass is $13.2 g$ .

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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