Question

Review I Constants Periodic Table The common-ion effect is an application of Le Châtelier's principle, which states that an equilibrium system that is stressed will work to alleviate that stress and reestablish equilibrium. Part B-Calculate the molar solubility in NaOH Based on the given value of the Ksp, what is the molar solubility of Mg(OH)2 in 0.190 M NaOH? Express your answer with the appropriate units. View Available Hint(s) molar solubilityValue Units Submit X Incorrect; Try Again Part C Complete previous part(s) The common-ion effect and bufter systems

Answer 1

Part B

Let the solubility of Mg(OH)₂ is S mol/L

it dissociates as, Mg(OH)₂    $\rightleftharpoons$   Mg²⁺   + 2OH^-

[Mg²⁺ ]= S mol/L

[OH^-] = 2S mol/L

Ksp of of Mg(OH)₂ is 1.8×10^–11

Ksp = [Mg²⁺ ][OH^-]²

Contribution from NaOH,

[OH^-] = 0.190 M or mol/L

Thus, overall [OH^-] = (0.190 + S) mol/L

Ksp = [Mg²⁺ ][OH^-]²

or, 1.8×10^–11 = (S)(0.190 + S)²

owing to very small Ksp value, S can be neglected w.r.t. 0.190

Hence,

1.8×10^–11 = (S)(0.190 )²

or, S = 4.98 x 10^-10 mol/L

$\therefore$ solubility of Mg(OH)₂ is 4.98 x 10^-10 mol/L