Please help by Monday
Explain in words how you would prepare 250.0 mL of a 3.000 M solution of Ba(OH)2 starting from solid barium hydroxide and water.
To prepare a 3.000 M solution of Ba(OH)2, you will need solid barium hydroxide and water. Here's how you can do it:
Step 1: Calculate the amount of barium hydroxide needed.
Since you want a final volume of 250.0 mL and a concentration of 3.000 M, you need to calculate the moles of barium hydroxide required. The formula for barium hydroxide is Ba(OH)2 and it has a molar mass of 171.34 g/mol.
The formula to calculate moles is:
moles = concentration (in M) x volume (in L)
moles = 3.000 M x 0.2500 L
moles = 0.7500 mol
Step 2: Convert moles to grams.
To convert moles to grams, you can use the molar mass. Multiply the moles of barium hydroxide by its molar mass.
grams = moles x molar mass
grams = 0.7500 mol x 171.34 g/mol
grams = 128.51 g
So, you will need 128.51 grams of solid barium hydroxide.
Step 3: Dissolving barium hydroxide in water.
Take a clean, dry beaker and add about 200 mL of deionized or distilled water. Use a weigh boat to accurately measure 128.51 grams of solid barium hydroxide. Slowly add the solid barium hydroxide to the water while gently stirring with a glass rod. Be cautious as the reaction is exothermic and may heat up. Continue to stir until all the solid barium hydroxide has dissolved.
Step 4: Adjusting the final volume.
Once all the solid barium hydroxide has dissolved, add more deionized or distilled water to the beaker to bring the total volume to 250.0 mL. Make sure to mix the solution well to ensure homogeneity.
And there you have it! You have prepared 250.0 mL of a 3.000 M solution of Barium Hydroxide (Ba(OH)2) starting from solid barium hydroxide and water.