What is the minimum volume of 2.00 M HCl necessary to neutralize completely the hydroxide in 800.0 mL of 0.3 M NaOH?
mols NaOH = M x L = ?
mols HCl = mols NaOH (look at the coefficients in the balanced equation).
M HCl = mols HCl/L HCl. You know mols ad M, solve for L (convert to mL if you wish).
To find the minimum volume of 2.00 M HCl necessary to completely neutralize the hydroxide in 800.0 mL of 0.3 M NaOH, we can use the concept of mole-to-mole ratios.
The balanced chemical equation for this reaction is:
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that one mole of NaOH reacts with one mole of HCl to produce one mole of water.
First, we need to determine the number of moles of NaOH present in 800.0 mL of 0.3 M NaOH.
Moles of NaOH = (Concentration of NaOH) x (Volume of NaOH in liters)
= 0.3 mol/L x 0.800 L
= 0.24 mol
Since the mole-to-mole ratio between NaOH and HCl is 1:1, we can determine that we need 0.24 moles of HCl to completely neutralize the NaOH.
Now, we need to find the volume of 2.00 M HCl required to provide 0.24 moles of HCl.
Volume of HCl = (Moles of HCl) / (Concentration of HCl)
= 0.24 mol / 2.00 mol/L
= 0.12 L or 120 mL
Therefore, the minimum volume of 2.00 M HCl necessary to neutralize completely the hydroxide in 800.0 mL of 0.3 M NaOH is 120 mL.