Calculate the volume of 0.500 M KOH required to neutralize 150 mL of 0.100 M HCl.
Group of answer choices
150 mL
750 mL
75 mL
30. mL
none of these
To find the volume of 0.500 M KOH required to neutralize 150 mL of 0.100 M HCl, we can use the concept of stoichiometry.
First, let's write the balanced chemical equation for the reaction between KOH and HCl:
KOH + HCl -> KCl + H2O
From the balanced equation, we can see that the stoichiometric ratio between KOH and HCl is 1:1. This means that 1 mole of KOH reacts with 1 mole of HCl.
Next, let's convert the given volumes of solutions into moles using the equation:
moles = concentration (M) × volume (L)
For the given HCl solution:
moles of HCl = 0.100 M × 0.150 L = 0.015 moles
Since the stoichiometric ratio is 1:1, this means that 0.015 moles of KOH will react with 0.015 moles of HCl.
Finally, let's find the volume of 0.500 M KOH required to contain 0.015 moles using the equation:
volume (L) = moles / concentration (M)
volume of KOH = 0.015 moles / 0.500 M = 0.030 L = 30 mL
Therefore, the volume of 0.500 M KOH required to neutralize 150 mL of 0.100 M HCl is 30 mL.