How much 0.300 M NaOH is required to completely neutralize 30.0 ml of 0.15 M HClO4
Clue: M1V1=M2V2
The hint works the problem for you. Substitute the numbers and solve.
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
See above.
To find out how much 0.300 M NaOH is required to completely neutralize 30.0 ml of 0.15 M HClO4, we can use the principle of stoichiometry and the given information.
The chemical equation for the reaction between NaOH and HClO4 is:
NaOH + HClO4 -> NaClO4 + H2O
The balanced equation tells us that one mole of NaOH reacts with one mole of HClO4. Therefore, the stoichiometric ratio between NaOH and HClO4 is 1:1.
Using the equation M1V1 = M2V2, where M refers to the molarity and V refers to the volume, we can calculate the amount of NaOH needed.
Given:
M1 (concentration of NaOH) = 0.300 M
V1 (volume of NaOH) = unknown
M2 (concentration of HClO4) = 0.15 M
V2 (volume of HClO4) = 30.0 ml
Since the stoichiometric ratio is 1:1, the moles of NaOH required will be equal to the moles of HClO4 present.
First, we need to convert the volume of HClO4 from milliliters (ml) to liters (L):
V2 = 30.0 ml = 30.0 ml / 1000 ml/L = 0.030 L
Using the equation M1V1 = M2V2 and substituting the known values:
0.300 M (NaOH) * V1 = 0.15 M (HClO4) * 0.030 L
Solving for V1 (volume of NaOH):
V1 = (0.15 M * 0.030 L) / 0.300 M
V1 = 0.015 L
Converting the volume of NaOH from liters (L) to milliliters (ml):
V1 = 0.015 L * 1000 ml/L = 15 ml
Therefore, 15.0 ml of 0.300 M NaOH is required to completely neutralize 30.0 ml of 0.15 M HClO4.