What volume of a 0.500 M NaOH solution is needed to neutralize 60.0 mL of 0.15 M HCl solution?
mols HCl = M x L = ?
mols NaOH = mols HCl (look at tahe coefficients; the rxn is 1:1).
Then M NaOH = mols NaOH/L NaOH.
To solve this problem, we need to use the concept of stoichiometry. Let's break it down step by step:
Step 1: Write the balanced chemical equation for the reaction between NaOH (sodium hydroxide) and HCl (hydrochloric acid):
NaOH + HCl -> NaCl + H2O
From the equation, we can see that one mole of NaOH reacts with one mole of HCl to produce one mole of NaCl and one mole of water.
Step 2: Determine the number of moles of HCl in the given volume of solution.
Moles of HCl = Concentration of HCl x Volume of HCl
= 0.15 mol/L x 0.060 L
= 0.009 mol
Step 3: Use the stoichiometry of the balanced equation to determine the number of moles of NaOH required to neutralize the HCl.
According to the balanced equation, one mole of NaOH reacts with one mole of HCl. Therefore, the number of moles of NaOH required is also 0.009 mol.
Step 4: Calculate the volume of the 0.500 M NaOH solution needed to have 0.009 moles.
Volume of NaOH = Moles of NaOH / Concentration of NaOH
= 0.009 mol / 0.500 mol/L
= 0.018 L or 18 mL
Therefore, 18 mL of the 0.500 M NaOH solution is needed to neutralize 60.0 mL of the 0.15 M HCl solution.