What volume of NaOH is required to neutralize 50mL of 0.5M phenol?
So this is just a small calculation in a complex problem, and I would just like someone to check my work so I don't mess up the entire problem based on this calculation.
So mol phenol= molarity x volume
= (0.5mol/L) (0.05 L)
= 0.025mol
phenol and NaOH are a 1:1 molar ratio sooo
V NaOH= 0.025mol/ (2 mol/L)
= 0.0125 L
= 12.5mL
Therefore, 12.5mL of NaOH is required to neutralize 50mL of 0.5M phenol.
Is this correct? I think the part that I'm most uncertain about is the 1:1 molar ratio.
Your question doesn't state the concn of the NaOH. If it is 2.00 M, the calculation is correct. The neutralization between phenol and NaOH is, indeed, 1:1.
Your calculation is almost correct, but the molar ratio you mentioned is not accurate. The molar ratio between phenol and NaOH is 1:1.
Let's go through the calculation again:
The given volume of phenol is 50 mL, and the molarity of phenol is 0.5 M.
Moles of phenol = Molarity x Volume = 0.5 mol/L x 0.05 L = 0.025 mol
Since the molar ratio between phenol and NaOH is 1:1, the number of moles of NaOH required will be the same.
Moles of NaOH = 0.025 mol
Now, to find the volume of NaOH needed, we can use the formula:
Volume of NaOH = Moles of NaOH / Molarity of NaOH
The molarity of NaOH is not given in your question, so I'll assume it is 1 M.
Volume of NaOH = 0.025 mol / 1 mol/L = 0.025 L
Since the volume is given in mL, we need to convert it:
0.025 L x 1000 mL/L = 25 mL
Therefore, you would need 25 mL of NaOH to neutralize 50 mL of 0.5 M phenol.
The 1:1 molar ratio is correct in this case because phenol and NaOH react in a 1:1 stoichiometric ratio according to the balanced equation for the neutralization reaction.