I don't remember ever doing a problem like this, so I am not sure about my answer, so hopefully someone comes along and fixes my mistakes if I made any.
0.2445 g*(1 mole/106 g of Na2CO3)=2.307 x 10^-3 moles of Na2CO3
The reaction shows 2HClO4 are needed for 1 Na2CO3, so 2*(2.307 x 10^-3)= moles of HClO4 reacted, but we have excess and we will let the unreacted excess =x.
M1=acid
M2=base
M1V2=M1V2
And we know in the final titration the volumes used, so plug in the volumes
M1*0.025L=M2*0.02688L
And the information from the back titration is as followed:
M1*Y=M2*0.00413L
Let Y= volume of unreacted volume of HClO4
Since the Molarity of the bases used in both titrations are the same, and the Molarity of the acid is the same in both titrations, then I can solve for Y
Y/0.025L=0.00413L/0.02688L, solving for Y
Y=0.025L*(0.00413L/0.02688L)=0.00384L
We have two equations, but we don't know the Molarity.
M1*0.025L=M2*0.02688L
M1*0.00384L=M2*0.00413L
Moles=Molarity/volume and the first titration is moles of unreacted HClO3, so
x=M*0.00384L
We know the Molarity of the original solution is:
4.614 x 10^-3moles+ x/0.050L=M
Substituting one equation into another:
(4.614 x 10^-3moles+ M1*0.00384L)/0.050L=M1
0.09228 moles/L + 0.0768 M=M
Solving for M,
0.09228 moles/L=0.9232 M
0.09996 moles/L=M
Since M1=0.09996
M2=0.09996(0.025L /0.02688L)=0.09297 M
0.004998 moles of Acid in 50 mL
3.838 x 10^-4 moles of acid in 25 mL
A dilute perchloric acid solution was standardized by dissolving 0.2445 g of primary standard sodium carbonate in 50 mL of the acid, boiling to eliminate CO2, and back-titrating with 4.13mL of dilute NaOH. In a separate titration a 25mL portion of the acid required 26.88 mL of the base. Calculate the molar concentrations of the acid and the base.
First i wrote the equations
Na2CO3 + 2HClO4 -> CO2 + 2NaClO4 + H2O
n(Na2CO3)=n(HClO4)/2
back titration
HClO4 + OH -> ClO4- + H2O
n(HClO4)=n(OH)=c(OH)*0.00413L
from the separate titration:
n(HClO4)=n(NaOH)
c(HClO4)*0.025L=c(NaOH)*0.02688L
from first equations i calculated the mols:
n(Na2CO3)=0.2445/105.99=0.0023068 mol
n(HClO4)=2*0.0023068=0.0046134 mol
i know that the mols from back titration represent the excess acid, but i just get stuck with the calculations.
4 answers
I agree with all of the numbers Devron but but I think there is an easier way to do it.
mol Na2CO3 = 0.2445/106 = 0.002307 which is equivalent to 2*0.002307 mols HClO4 = 0.04614 mols HClO4. Now all we need to do is find volume HClO4. We know it is 50.0 initially. How much extra HClO4 was present? That's
4.13 mL NaOH x (25.0 mL HClO4/26.88 mL NaOH) = 3.84 mL HClO4.
Therefore, the volume HClO4 was 50.0-3.84 = 46.16 mL and
M HClO4 = 0.04614/0.04616 = 0.09996
M NaOH = 0.09996*25/26.88 = 0.09297
I like to carry an extra place in ALL of the calculations and round at the very end. That sometimes makes a difference of 1 or 2 in the last digits.
mol Na2CO3 = 0.2445/106 = 0.002307 which is equivalent to 2*0.002307 mols HClO4 = 0.04614 mols HClO4. Now all we need to do is find volume HClO4. We know it is 50.0 initially. How much extra HClO4 was present? That's
4.13 mL NaOH x (25.0 mL HClO4/26.88 mL NaOH) = 3.84 mL HClO4.
Therefore, the volume HClO4 was 50.0-3.84 = 46.16 mL and
M HClO4 = 0.04614/0.04616 = 0.09996
M NaOH = 0.09996*25/26.88 = 0.09297
I like to carry an extra place in ALL of the calculations and round at the very end. That sometimes makes a difference of 1 or 2 in the last digits.
I thought it was an easier way to do it, but it was super early/late and I couldn't reason it at the time.
Thank you both so much
1 answer