How much 0.300 M NaOH is required to completely neutralize 30.0 ml of 0.15 M HClO4
Clue: M1V1=M2V2
I still don't get it to i times .300x30.0 then times .300 x.15 i just don't understand what i am actually suppose to multiply
M NaOH x mL NaOH = M HClO4 x mL HClO4
0.300M x mL NaOH = 0.15M x 30.0 mL
You have one unknown; i.e., mL NaOH. Solve for that.
To determine the amount of 0.300 M NaOH required to neutralize 30.0 mL of 0.15 M HClO4, you can use the equation M1V1 = M2V2.
Let's break down the equation:
M1 represents the concentration of the first solution (HClO4) in moles per liter (Molarity).
V1 represents the volume of the first solution (HClO4) in liters.
M2 represents the concentration of the second solution (NaOH) in moles per liter (Molarity).
V2 represents the volume of the second solution (NaOH) in liters.
In this case:
M1 = 0.15 M (HClO4 concentration)
V1 = 30.0 mL (HClO4 volume)
M2 = 0.300 M (NaOH concentration)
V2 = ? (NaOH volume, what we are trying to find)
Now, let's substitute the known values into the equation:
0.15 M × 30.0 mL = 0.300 M × V2
To solve for V2 (NaOH volume), you need to isolate it on one side of the equation.
Divide both sides of the equation by 0.300 M:
(0.15 M × 30.0 mL) / 0.300 M = V2
Now, calculate the answer:
V2 = 15.0 mL
Therefore, 15.0 mL of 0.300 M NaOH is required to completely neutralize 30.0 mL of 0.15 M HClO4.