Final answer:
To neutralize 100 mL of a 6M HCl solution, you would need 13.8 grams of NaOH.
Step-by-step explanation:
To find the number of grams of NaOH required to neutralize 100 mL of a 6M HCl solution, we can use the equation:
HCl + NaOH → NaCl + H₂O
The balanced equation tells us that the ratio of moles of HCl to NaOH is 1:1. Given that the concentration of the HCl solution is 6M, we can calculate the number of moles of HCl in 100 mL (0.1 L) using the formula:
Moles of solute (HCl) = concentration (M) x volume (L)
Substituting the values, we get:
Moles of HCl = 6 mol/L x 0.1 L = 0.6 mol HCl
Since the ratio of moles of HCl to NaOH is 1:1, we need 0.6 moles of NaOH to neutralize the HCl. The molar mass of NaOH is 23 g/mol, so the mass of NaOH required can be calculated as:
Mass of NaOH = moles of NaOH x molar mass of NaOH = 0.6 mol x 23 g/mol = 13.8 g NaOH