Final answer:
The concentration of OH- when 25 grams of Al(OH)3 is added to 2.50 L of water is 0.13 M.
Step-by-step explanation:
To find the concentration of OH- when 25 grams of Al(OH)3 is added to 2.50 L of water, we need to calculate the number of moles of Al(OH)3 and then use the volume of water to find the concentration.
First, calculate the number of moles of Al(OH)3 using the molar mass:
moles of Al(OH)3 = (mass of Al(OH)3)/(molar mass of Al(OH)3) = (25 g)/(78.00 g/mol) = 0.32 mol
Next, calculate the concentration using:
concentration = moles/volume = 0.32 mol/2.50 L = 0.13 M