NEETChemistryIonic Equilibrium
Calculate the solubility of Mg(OH) ₂ in a buffer solution of pH 12. Given that the solubility product ( K_ sp ) of Mg(OH) ₂ is 1 10⁻¹¹ .
Options
- A1 10⁻⁹ M
- B2.5 10⁻⁸ M
- C1 10⁻⁷ M
- D1 10¹³ M
Correct answer
C. 1 10⁻⁷ M
Step-by-step solution
Given pH of the buffer solution = 12. We know that pH + pOH = 14 pOH = 14 - 12 = 2 Therefore, the concentration of hydroxide ions is: [ OH ^-] = 10^ - pOH = 10⁻² M The dissociation of Mg(OH) ₂ is given by: Mg(OH) ₂(s) Mg ²⁺(aq) + 2 OH ^-(aq) Let the solubility of Mg(OH) ₂ in the buffer be s . Then, [ Mg ²⁺] = s . Due to the buffer, the [ OH ^-] is maintained at 10⁻² M . The solubility product expression is: K_ sp = [ Mg ²⁺][ OH ^-]^2 Substitute the given values into the expression: 1 10⁻¹¹ = s (10⁻²)^2 1 10⁻¹¹ = s