"How would you prepare 150. mL of 0.1M sodium hydroxide given a stock solution of 3.0M NaOH?"
I made calculations and figured the process would be to dilute 5 mL of the 3.0M NaOH to 150. mL. Is this correct?
For the ideal 150. mL of 0.1M NaOH, I calculated there to need .015 mol of NaOH in the solution, which is present in 5 mL of 3.0M NaOH.
You're right. Go to the head of the class.
3.0M in V ml
0.1M in 150 mL
dilution factor: 30x
Therefore, 150/30 = 5ml from the stock solution
To correctly prepare 150 mL of 0.1M sodium hydroxide (NaOH) solution using a stock solution of 3.0M NaOH, you need to follow the process of dilution accurately.
First, let's determine the amount of NaOH (in moles) required to make a 0.1M solution. The concentration of the desired solution is 0.1M, and the volume is 150 mL. We can use the formula:
moles = concentration * volume
moles = 0.1 M * 0.15 L = 0.015 mol
Now, you correctly found that you need 0.015 mol of NaOH in the final solution.
To calculate the volume of the stock solution you need to dilute, you can rearrange the formula mentioned earlier:
volume of stock solution = moles / concentration
volume of stock solution = 0.015 mol / 3.0 M = 0.005 L (or 5 mL)
So, to prepare 150 mL of 0.1M NaOH solution, you should take 5 mL of the 3.0M NaOH stock solution and then add enough water to reach the final volume of 150 mL. This would ensure the correct concentration is achieved.
Therefore, your initial calculation of taking 5 mL of the 3.0M NaOH stock solution to prepare 150 mL of 0.1M NaOH solution is indeed correct. Just remember to make up the remaining volume with water.