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The minimum conc. of NH 4 Cl required to prevent the precipitation of Mg(OH) 2 from a solution of 0.05 M Mg 2+ and 0.05 M NH 3 is x 10 -3 . What is the value of x in nearest whole number ? (K sp ( Mg(OH)2) = 9.0 x 10 -12 , K b (NH 4 OH) = 1.8 x 10 -5 )

Options

  1. A67
  2. B51
  3. C16
  4. D8

Correct answer

A. 67

Step-by-step solution

Conc. of OH - = K sp Mg 2 + = 1.34 × 1 0 - 5 K NH 4 OH = NH 4 + 1.34 × 1 0 - 5 0.05 NH 4 + = 0.0671 M = 67.1 × 1 0 - 3 M So, answer is 67. Alternative solution IF Mg(OH) 2(s) is just formed then the equilibria existing are Mg OH 2 s + Aq. ⇌ Mg 2 + + 2 OH aq - ; K sp 2 OH aq - + 2 NH 4 aq + ⇌ 2 NH 3 + 2 H 2 O ; 1 K b 2 Mg OH 2 s + 2 NH 4 aq + ⇌ Mg 2 + + 2 NH 3 + 2 H 2 O ; ; K eq = K sp K b 2 K eq = 9 × 1 0 - 1 2 1.8 × 1 0 - 5 2 = Mg 2 + NH 3 2 NH 4 + 2 NH 4 + 2 = Mg 2 + NH 3 2 9 ×

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