What is the hydroxide ion concentration of a solution prepared by dissolving 0.488 g of Mg(OH)2 in 236 mL of water? How do I solve this???

mols Mg(OH)2 = grams/molar mass

M = mols/L soln. ( I assume the 236 mL of water is 236 mL solution.) I obtained M = about 0.04M but you need to do it more accurately. I'm also assuming that all of the 0.488 g dissolves.
M = about 0.04M
There are 2 OH^-/1 molecule Mg(OH)2; therefore, (OH^-) = twice that or about 0.08M

To solve this question, we need to calculate the hydroxide ion concentration of the solution. Let's break it down step-by-step:

1. Start by converting the mass of Mg(OH)2 to moles.
- The molar mass of Mg(OH)2 is calculated as:
(1 mol of Mg x molar mass of Mg) + (2 mol of O x molar mass of O) + (2 mol of H x molar mass of H)
= 24.31 g/mol + 15.999 g/mol + (2 x 1.0078 g/mol) = 58.3192 g/mol

- Divide the mass of Mg(OH)2 by its molar mass to get the number of moles:
0.488 g / 58.3192 g/mol = 0.00838 mol

2. Next, calculate the number of moles in the solution.
- The volume of the solution is given as 236 mL, so we need to convert it to liters by dividing by 1000:
236 mL / 1000 mL/L = 0.236 L

3. Calculate the hydroxide ion concentration (OH-) in moles per liter (M).
- The hydroxide ion concentration is determined by dividing the number of moles of Mg(OH)2 by the volume of the solution in liters:
Concentration of OH- = 0.00838 mol / 0.236 L = 0.0355 M

So, the hydroxide ion concentration of the solution prepared by dissolving 0.488 g of Mg(OH)2 in 236 mL of water is 0.0355 M.