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Mg OH 2 is precipitated when NaOH is added to a solution of Mg 2 + . If the final concentration of Mg 2 + is 10 - 10   M , the concentration of OH - M in the solution is [Solubility product for Mg OH 2 = 5 . 6 × 10 - 12 ]

Options

  1. A0 . 056
  2. B0 . 12
  3. C0 . 24
  4. D0 . 025

Correct answer

C. 0 . 24

Step-by-step solution

In the saturated solution of sparingly soluble electrolyte, the ionic product of ions is constant at constant temperature and it is known a Solubility Product. Here, Mg ( OH ) 2 → Mg 2 + OH - 2 Given, K sp for Mg ( OH ) 2 = 5 . 6 × 10 - 12 Mg 2 + = 10 - 10   M Substituting the value, we get : K sp = Mg 2 + OH - 2 ⇒ 5 . 6 × 10 - 12 = 10 - 10 OH - 2 ⇒ OH - = 0 . 24   M Hence, the concentration of [ OH - ] = 0 . 24   M

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