Final answer:
To separate benzoic acid and 2-napthol, a strong base can be added to react with benzoic acid and form a soluble salt, while leaving 2-napthol unchanged. To neutralize a 2.0 M NaOH solution, 20 mL of 6.0 M HCl solution is needed.
Step-by-step explanation:
In an acid-base extraction, benzoic acid and 2-napthol can be separated based on their different acid strengths. Since benzoic acid is a stronger acid than 2-napthol, it will react with a strong base to form its conjugate base, while 2-napthol will remain mostly unchanged.
By adding a strong base, such as sodium hydroxide (NaOH), to the mixture, the benzoic acid will react and form a soluble sodium salt, while 2-napthol will remain in its neutral form and can be separated.
To completely neutralize a 2.0 M NaOH solution, we can use the concept of stoichiometry. The balanced equation for the neutralization reaction between HCl and NaOH is: NaOH + HCl -> NaCl + H2O. From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of water. Therefore, to neutralize 60 mL of 2.0 M NaOH solution, we need to calculate the amount of moles of NaOH first:
Moles of NaOH = Volume (L) × Molarity = 0.060 L × 2.0 M = 0.120 mol NaOH
Since the reaction is 1:1 between NaOH and HCl, we need the same amount of moles of HCl to completely neutralize the NaOH. Now we can calculate the amount of volume of 6.0 M HCl solution needed:
Volume of HCl = Moles / Molarity = 0.120 mol / 6.0 M = 0.020 L = 20 mL