What is the minimum volume of a 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution?
To determine the minimum volume of a 2.00 M NaOH solution needed to make a 0.800 M NaOH solution, we can use the equation: M1V1 = M2V2.
Where:
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
Let's substitute the given values into the equation to solve for V1:
M1 = 2.00 M
V1 = unknown
M2 = 0.800 M
V2 = 150.0 mL (or 0.150 L)
2.00 M * V1 = 0.800 M * 0.150 L
Now we can solve for V1:
V1 = (0.800 M * 0.150 L) / 2.00 M
V1 = 0.060 L
Therefore, the minimum volume of the 2.00 M NaOH solution needed to make 150.0 mL of a 0.800 M NaOH solution is 0.060 L (or 60.0 mL).