A stock solution was prepared by dissolving exactly 0.4000 g of pure ASA (180.16 g/mol)
in 10.00 mL of NaOH and heating the solution to a gentle boil. After cooling to room
temperature, the solution was poured into a 250-mL volumetric flask and diluted to the mark
with DI water. Before coming to lab, calculate the concentration of ASA in the flask.
I got my molarity as 0.00888M:
0.400g/180.16=0.00222mols/0.250= 0.00888M
however am I suppose to take into account the 10.0ml of NaOH and if so how would I do that?
Your calculated molarity is correct. No, you don't need to "correct" for the 10 mL NaOH because that 10 mL is part of the solvent used when made to the 250 mL mark. That is, the 10 mL is part of the 250 mL.
Okay thank you that makes a lot more sense!
To calculate the concentration of ASA in the flask, you need to consider the volume of NaOH added to the solution. From the question, it is stated that 10.00 mL of NaOH was used.
To calculate the concentration of ASA taking into account the volume of NaOH, you can follow these steps:
Step 1: Calculate the molarity of NaOH in the solution.
In this case, since the volume of NaOH used is 10.00 mL, we'll assume it is a 1 M solution.
Therefore, the concentration of NaOH is:
1 M = 1 mole of NaOH / 1 L of solution
So, for the 10.00 mL of NaOH:
Concentration of NaOH = (1 mole / 1 L) x (10.00 mL / 1000 mL) = 0.01 moles / 0.01 L = 1 M
Step 2: Calculate the moles of ASA in the solution.
You have correctly calculated the moles of ASA as 0.00222 moles.
Step 3: Adjust the moles of ASA to account for the dilution caused by adding the NaOH solution.
Since the ASA was dissolved in 10.00 mL of NaOH, the total volume of the solution is 10.00 mL.
However, after diluting to a 250 mL volumetric flask, the final volume is 250 mL.
The moles of ASA in the final solution can be calculated using the following proportion:
Moles of ASA in the final solution = (0.00222 moles / 10.00 mL) x 250 mL
= 0.00222 moles x (250 mL / 10.00 mL)
= 0.00222 moles x 25
= 0.0555 moles
Step 4: Calculate the concentration of ASA in the flask.
To calculate the concentration of ASA in the flask, divide the moles of ASA in the final solution by the total volume in liters (converted from mL):
Concentration of ASA = 0.0555 moles / (250 mL / 1000 mL/L)
= 0.0555 moles / 0.25 L
= 0.222 M
Therefore, the concentration of ASA in the flask is 0.222 M.
To calculate the concentration of ASA in the volumetric flask, you need to take into account the volume of both the ASA solution and the NaOH solution.
First, calculate the number of moles of ASA in the 0.4000 g of pure ASA using the molar mass of ASA.
moles of ASA = mass of ASA / molar mass of ASA
moles of ASA = 0.4000 g / 180.16 g/mol
moles of ASA = 0.00222 mol
Next, calculate the total volume of the solution after dilution. The initial volume of the ASA solution (10.00 mL) should be added to the final volume in the volumetric flask (250 mL).
total volume of the solution = initial volume of ASA solution + final volume in the volumetric flask
total volume of the solution = 10.00 mL + 250 mL
total volume of the solution = 260 mL = 0.260 L
Now, divide the moles of ASA by the total volume of the solution to find the concentration of ASA in moles per liter (Molarity).
concentration of ASA = moles of ASA / total volume of the solution
concentration of ASA = 0.00222 mol / 0.260 L
concentration of ASA = 0.00854 M
So, the correct concentration of ASA in the volumetric flask is approximately 0.00854 M.