How much 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4?

Solution 90P :

Step 1:

Given :

Molarity of KOH(Potassium hydroxide)(M1) = 0.200 M

Molarity of HClO4(Perchloric acid) (M2) = 0.150 M

Volume of HClO4 (V2)= 25.0 mL

Volume of KOH to neutralize HClO4 (V1) = ?

Let’s write the chemical reaction for the given scenario :

KOH + HClO4 → KClO4 + H2O

We know the formula to calculate the molarity of the diluted solution, :

M1V1 = M2V2

Where, M1 and V1 = Molarity and volume of initial concentrated solution.

         M2 and V2 = Molarity and volume of final diluted solution.  

Using the above formula we calculate the volume of KOH to neutralize HClO4

We have to calculate V1, hence the formula becomes :

                V1 =

Volume of HClO4 is in mL, so let’s convert to L :

                1 L = 1000 mL

Therefore, 25.0 mL in L will be :

                = 25 mL

                = 0.025 L