25.00 mL of a solution of oxalic acid are titrated with 0.2586 m NaOH (aq). The stoichiometric eend point is reached when 43.42 mL of the solution of base is added. What is the molarity of the oxalic acid solution? Oxalic acid reacts with sodium hydroxide as shown:
H2C2O4(aq) + 2 NaOH(aq) --> Na2C2O4(aq) + 2 H2O (l)
I thought I had this figured out last night, but now as many times as I try I can only come up with .044
I had .01123
And I don't think 0.044 is it.
I would call your attention to the 0.2586m NaOH. If that should be 0.2586 M and I assume it should, you should be aware that m stands for molality while M stands for molarity. They are two different animals. A word to the wise?
mols NaOH = M x L = 0.2586M x 0.04342L = about .011.
Use the coefficients in the balanced equation to convert mols NaOH to mols H2C2O4.
mols H2C2O4 = 1/2 mol NaOH = about 0.005.
M H2C2O4 = mols H2C2O4/L H2C2O4 = about 0.005/0.025 = about 0.22M
Thank you!
Yes, I meant M, sorry :)
To determine the molarity of the oxalic acid solution, you need to use the balanced equation for the reaction and the volume of the NaOH solution.
First, let's calculate the number of moles of NaOH used in the titration:
Moles of NaOH = concentration of NaOH × volume of NaOH solution
Moles of NaOH = 0.2586 M × 43.42 mL = 11.2406 mmol (millimoles)
Now, since the stoichiometric ratio between NaOH and oxalic acid is 2:1, the number of moles of oxalic acid is half the number of moles of NaOH used:
Moles of oxalic acid = 11.2406 mmol / 2 = 5.6203 mmol
To determine the molarity of the oxalic acid solution, divide the moles of oxalic acid by the volume of the oxalic acid solution (in liters):
Molarity of oxalic acid = Moles of oxalic acid / Volume of oxalic acid solution (in liters)
Molarity of oxalic acid = 5.6203 mmol / 25.00 mL / 1000 = 0.2248 M
Therefore, the molarity of the oxalic acid solution is 0.2248 M. Note that this calculation assumes that the titration reached the stoichiometric endpoint accurately.
To determine the molarity of the oxalic acid solution, you need to use the stoichiometry of the balanced chemical equation and the volume of NaOH solution required to reach the stoichiometric endpoint.
First, calculate the moles of NaOH used in the titration:
moles of NaOH = molarity of NaOH × volume of NaOH solution
moles of NaOH = 0.2586 M × 43.42 mL
Next, use the stoichiometric ratio between NaOH and H2C2O4 to find the moles of oxalic acid:
moles of H2C2O4 = (moles of NaOH) × (1 mol H2C2O4 / 2 mol NaOH)
Since the volume of the oxalic acid solution is given in mL, you must convert it to liters to calculate the molarity:
volume of oxalic acid solution (L) = 25.00 mL × (1 L / 1000 mL)
Finally, calculate the molarity of the oxalic acid solution:
molarity of H2C2O4 = (moles of H2C2O4) / (volume of H2C2O4 solution)
Let's now plug in the values and calculate the molarity:
moles of NaOH = 0.2586 M × 43.42 mL = 0.01123612 mol NaOH
moles of H2C2O4 = (0.01123612 mol NaOH) × (1 mol H2C2O4 / 2 mol NaOH) = 0.00561806 mol H2C2O4
volume of H2C2O4 solution = 25.00 mL × (1 L / 1000 mL) = 0.02500 L
molarity of H2C2O4 = (0.00561806 mol H2C2O4) / (0.02500 L) = 0.2247 M
Therefore, the molarity of the oxalic acid solution is 0.2247 M.