Final answer:
The molar solubility of Mg(OH)2 in pure water is approximately 1.4×10^-4 M, calculated using the Ksp value of 5.61×10^-11. In 0.100 M NaOH, the solubility of Mg(OH)2 would be lower due to the common-ion effect.
Step-by-step explanation:
The molar solubility of Mg(OH)2 in pure H2O can be calculated using its Ksp value of 5.61×10−11. The dissolution of Mg(OH)2 can be written as: Mg(OH)2 (s) ⇒ Mg2+ (aq) + 2OH− (aq). Using the Ksp expression Ksp = [Mg2+][OH−]2, and letting the molar solubility be 's', we have 5.61×10−11 = s(s×2)2. Solving for 's', we get the molar solubility is approximately 1.4×10−4 M.
When Mg(OH)2 is in a solution with 0.100 M NaOH, the increased concentration of OH− ions from the NaOH affects the solubility. The common-ion effect will decrease the solubility of Mg(OH)2 since the product of [Mg2+] and [OH−]2 must still equal the constant Ksp value. As [OH−] is already high, [Mg2+] must decrease, resulting in lower solubility of Mg(OH)2.
The molar solubility of Mg(OH)2 in pure H2O can be determined by using the solubility product constant, Ksp. In this case, the Ksp value is given as 5.61×10-11. The molar solubility can be calculated by taking the square root of the Ksp value. Therefore, the molar solubility of Mg(OH)2 in pure H2O is approximately 7.50×10-6 M.
The molar solubility of Mg(OH)2 in 0.100 M NaOH can be determined by considering the common ion effect. NaOH provides extra OH- ions, which can reduce the solubility of Mg(OH)2. In this case, the concentration of OH- ions is 0.100 M. By using the Ksp expression for Mg(OH)2 and considering the change in concentration of OH- ions, the molar solubility of Mg(OH)2 can be calculated. The molar solubility of Mg(OH)2 in 0.100 M NaOH is approximately 2.76×10-6 M.