Question

Describe how you would prepare \(750.0 \mathrm{~mL}\) of \(0.362 \mathrm{M}\) barium hydroxide solution starting with (a) solid barium hydroxide. (b) \(4.93 \mathrm{M}\) barium hydroxide solution.

Short answer

Answer: To prepare 750.0 mL of 0.362 M barium hydroxide solution, you will need approximately 46.53 grams of solid barium hydroxide and approximately 55.0 mL of the 4.93 M barium hydroxide solution.

Step-by-step solution

Calculate the number of moles of barium hydroxide needed for the solution

To find the moles of barium hydroxide needed, you can use the formula: moles = Molarity × Volume (in Liters). moles = 0.362 M × (750.0 mL / 1000) moles = 0.362 M × 0.750 L moles ≈ 0.2715 So, we need approximately 0.2715 moles of barium hydroxide.

Calculate the mass of solid barium hydroxide needed

To find the mass of barium hydroxide equivalent to 0.2715 moles, you can use the formula: Mass = moles × molar mass. The molar mass of barium hydroxide is 171.34 g/mol. Mass = 0.2715 moles × 171.34 g/mol ≈ 46.53 g So, we need approximately 46.53 grams of solid barium hydroxide.

Mix the solid barium hydroxide with water

To prepare the 0.362 M barium hydroxide solution, dissolve 46.53 grams of solid barium hydroxide in enough water to make a total volume of 750.0 mL. #For case (b):# In this case, we are given a 4.93 M barium hydroxide solution. To create a 0.362 M barium hydroxide solution from this, we first need to calculate the volume of the concentrated solution required.

Calculate the volume required of the 4.93 M barium hydroxide solution

To find the volume required from the 4.93 M barium hydroxide solution, you can use the formula: M1 × V1 = M2 × V2, where M1 and V1 are the molarity and volume of the concentrated solution and M2 and V2 are the molarity and volume of the diluted solution. 4.93 M × V1 = 0.362 M × 0.750 L V1 ≈ (0.362 M × 0.750 L) / 4.93 M V1 ≈ 0.0550 L or 55.0 mL So, we need approximately 55.0 mL of the 4.93 M barium hydroxide solution.

Dilute the concentrated solution

To prepare the 0.362 M barium hydroxide solution, mix 55.0 mL of the 4.93 M barium hydroxide solution with enough water to make a total volume of 750.0 mL.