The solubility product ( K_ sp ) of a sparingly soluble hydroxide X(OH) ₂ is 4 10⁻¹⁵ at 298 K . What is the solubility of X(OH) ₂ in a buffer solution maintained at pH = 10 ?

Step-by-step solution

Given pH = 10 , we have pOH = 14 - 10 = 4 . Thus, the hydroxide ion concentration is [ OH ^-] = 10⁻⁴ M . For the sparingly soluble salt X(OH) ₂ , the dissociation equilibrium is: X(OH) ₂ (s) X ²⁺ (aq) + 2 OH ^- (aq) The solubility product expression is: K_ sp = [ X ²⁺][ OH ^-]^2 Let the solubility of X(OH) ₂ in the buffer be s . Then [ X ²⁺] = s . Since the solution is buffered at pH = 10 , the [ OH ^-] is fixed at 10⁻⁴ M . Substituting the values into the K_ sp expression: 4 10⁻¹⁵ = s(10⁻⁴)^2 4 10⁻¹⁵ = s 10⁻⁸ s =