Base = NaOH, 0.084 M drips 2ml at a time
Acid = Phenol, KA=1.3*10^-10
You have 28.0 ml in flask, and it is 0.06 M
How would you find the molarity and moles of OC6H5-1
To find the molarity and moles of OC6H5-1 (phenolate ion), you can follow these steps:
Step 1: Write down the balanced chemical equation for the reaction between NaOH (base) and phenol (acid):
NaOH + OC6H5H → OC6H5-1 + H2O
Step 2: Determine the mole ratio between NaOH and OC6H5-1 in the balanced equation. From the equation, you can see that 1 mole of NaOH reacts with 1 mole of OC6H5-1.
Step 3: Calculate the moles of NaOH used in the reaction. To do this, use the volume of NaOH drips and the molarity of NaOH:
moles of NaOH = volume (L) x molarity (mol/L)
= (2 mL / 1000 mL/L) x (0.084 mol/L)
= 0.002 moles NaOH
Step 4: Since the mole ratio between NaOH and OC6H5-1 is 1:1, the moles of OC6H5-1 formed in the reaction will be equal to the moles of NaOH used:
moles of OC6H5-1 = moles of NaOH
= 0.002 moles
Step 5: Calculate the molarity of OC6H5-1 in the flask. To do this, use the volume and concentration of the flask solution:
moles of OC6H5-1 = molarity x volume (L)
0.002 moles = molarity x (28.0 mL / 1000 mL/L)
Rearranging the equation, we can solve for the molarity:
molarity = 0.002 moles / (28.0 mL / 1000 mL/L)
= 0.071 M
Therefore, the molarity of OC6H5-1 in the flask is 0.071 M and there are 0.002 moles of OC6H5-1 present.