What is the concentration of OH– ions formed when 40.0 g of NaOH is dissolved in 250 mL H2O?

To find the concentration of OH– ions formed when NaOH is dissolved in water, we need to use the concept of molarity. Molarity (M) is defined as the number of moles of solute per liter of solution.

First, we need to calculate the number of moles of NaOH using its molar mass. Sodium (Na) has a molar mass of 22.99 g/mol, oxygen (O) has a molar mass of 16.00 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol.

Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.99 g/mol

Next, we calculate the number of moles:
moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 40.0 g / 39.99 g/mol = 1.00 mol

Now we convert the volume of water from milliliters to liters:
volume of water = 250 mL = 250/1000 L = 0.250 L

Finally, we can calculate the concentration of OH– ions (NaOH dissociates into Na+ and OH– ions in solution):
concentration of OH– ions = moles of NaOH / volume of water
concentration of OH– ions = 1.00 mol / 0.250 L = 4.00 M

Therefore, the concentration of OH– ions formed when 40.0 g of NaOH is dissolved in 250 mL of water is 4.00 M.