Final answer:
To reduce the concentration of dissolved Mg2+ to 25 mg/L, a pH of approximately 17.134 is required.
Step-by-step explanation:
The concentration of Mg2+ can be determined using the solubility product constant (Ksp) for Mg(OH)2 and the given information. In a saturated solution of Mg(OH)2, the concentration of Mg2+ is given as 1.31 x 10-4 M. To find the solubility product constant (Ksp), we can set up the equilibrium equation:
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
From this equation, we can determine that the concentration of OH- ions is twice the concentration of Mg2+ ions. Therefore, the concentration of OH- ions is 2 x 1.31 x 10-4 M = 2.62 x 10-4 M.
Since the concentration of OH- ions is known, we can calculate the pH required to reduce the concentration of Mg2+ to 25 mg/L (which is the same as 25 x 10-3 g/L). The concentration of Mg2+ can be converted to moles per liter (mol/L) and compared to the concentration of OH- ions. From the balanced equation, we know that the ratio of Mg2+ ions to OH- ions is 1:2. Therefore, to reduce the concentration of Mg2+ to 25 mg/L, the concentration of OH- ions must be reduced to half that value, or 12.5 mg/L (or 12.5 x 10-3 g/L).
Next, we need to determine the molar concentration of OH- ions required to achieve this concentration in moles per liter (mol/L). We can use the molar mass of OH- (17.01 g/mol) to convert the given concentration from grams per liter (g/L) to moles per liter (mol/L):
12.5 x 10-3 g/L / 17.01 g/mol = 0.735 x 10-3 mol/L.
Now, we know that the concentration of OH- ions must be 0.735 x 10-3 mol/L. To calculate the pH required to achieve this concentration, we can use the equation:
pOH = -log[OH-]
Since pOH + pH = 14 (at 25°C), we can rearrange the equation to solve for pH:
pH = 14 - pOH
pH = 14 - (-log[OH-])
pH = 14 - (-log(0.735 x 10-3))
pH = 14 - (-3.134)
pH ≈ 17.134
Therefore, a pH of approximately 17.134 is required to reduce the concentration of dissolved Mg2+ to 25 mg/L.