Calculate the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M.
This is done the same way your other HCl/NaOH was done. You need to realize the 0.5 mol/L = 0.5M
To calculate the volume of NaOH 0.5 mol/L necessary to neutralize 300 mL of HCl 0.2 M, you need to use the equation:
M1 * V1 = M2 * V2
where M1 is the molarity of the first solution, V1 is the volume of the first solution, M2 is the molarity of the second solution, and V2 is the volume of the second solution.
In this case, HCl is the first solution with a molarity of 0.2 M and a volume of 300 mL. NaOH is the second solution with a molarity of 0.5 mol/L.
Let's plug the values into the equation:
(0.2 M) * (300 mL) = (0.5 mol/L) * (V2)
Now, let's calculate V2:
(0.2 M * 300 mL) / (0.5 mol/L) = V2
This calculation gives us the volume of NaOH required to neutralize the HCl solution.