When excess solid Mg(OH)2 is shaken with 1.00 L of 1.2 M NH4Cl solution, the resulting saturated solution has pH = 9.30. Calculate the Ksp of Mg(OH)2
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To calculate the Ksp of Mg(OH)2, we need to first write the balanced chemical equation for the dissociation of Mg(OH)2 in water:
Mg(OH)2 (s) ⇌ Mg2+ (aq) + 2 OH- (aq)
The solubility product expression for Mg(OH)2 is given by:
Ksp = [Mg2+] * [OH-]^2
Now, let's use the given information about the pH of the saturated solution to calculate the concentrations of Mg2+ and OH- ions.
Given: pH = 9.30
We know that pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration, so we can use it to find the hydroxide ion concentration.
pH = -log[H+] = 9.30
[H+] = 10^(-9.30)
Since water dissociates into H+ (hydrogen ions) and OH- (hydroxide ions), and we can assume that [H+] is equal to [OH-] in pure water, we can substitute [OH-] for [H+] in the solubility product expression.
Ksp = [Mg2+] * [OH-]^2
Now, we have to find the concentration of Mg2+ ions. To do that, we need to find the concentration of OH- ions.
[OH-] = [H+] = 10^(-9.30)
Since the balanced equation for the dissociation of Mg(OH)2 shows that there are two OH- ions produced for every Mg2+ ion, we can divide [OH-] by 2 to get the concentration of Mg2+ ions.
[Mg2+] = [OH-] / 2 = 10^(-9.30) / 2
Now, substitute the values into the Ksp expression:
Ksp = [Mg2+] * [OH-]^2 = (10^(-9.30) / 2) * (10^(-9.30))^2
Simplifying further:
Ksp = (10^(-9.30) / 2) * (10^(-18.60))
Ksp = 5 * 10^(-9.30 - 18.60)
Ksp = 5 * 10^(-27.90)
Finally, the Ksp of Mg(OH)2 is approximately 5 x 10^(-27.90).