What volume of 0.400 M HCl solution is needed to neutralize 30.0 ml of a 0.200 molar solution of KOH?
HCl + KOH ==> KCl + H2O
How many mols KOH? That's M x L = 0.200 x 0.030 = ?
The equation tells you 1 mol HCl needs 1 mol KOH; therefore, mols HCl = mols KOH.
Then mols KOH = M x L. You know M and mols, solve for L needed.
To answer this question, you need to use the concept of stoichiometry and the balanced chemical equation for the reaction between HCl and KOH. The balanced chemical equation is:
HCl + KOH -> KCl + H2O
Firstly, calculate the number of moles of KOH in the 30.0 ml (0.030 L) solution using the formula:
moles = molarity x volume
moles of KOH = 0.200 mol/L x 0.030 L = 0.006 mol
According to the balanced equation, the stoichiometry between HCl and KOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of KOH to produce 1 mole of KCl and 1 mole of H2O.
Since the stoichiometry is 1:1, the number of moles of HCl required to neutralize the given amount of KOH is also 0.006 mol.
Now, you need to find the volume of the 0.400 M HCl solution required to contain 0.006 moles of HCl. Rearranging the formula:
moles = molarity x volume
Volume of HCl = moles / molarity = 0.006 mol / 0.400 mol/L = 0.015 L or 15.0 ml
Therefore, 15.0 ml of the 0.400 M HCl solution is required to neutralize 30.0 ml of the 0.200 M KOH solution.