How much 0.200 M KOH is required to completely neutralize 25.0 mL of 0.150 M HClO4?
Solution 90P :
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 :
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