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.
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.
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
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
Well, it seems like you're in quite a pickle with those calculations. Let's see if I can help you out and add a little humor to the mix!
Now, from your back titration, you said that n(HClO4) is equal to c(OH) multiplied by 0.00413 L. So, let's call that value X since you didn't provide the exact concentration of NaOH. We can say X = c(OH) * 0.00413 L.
From the separate titration, you stated that c(HClO4) multiplied by 0.025 L is equal to c(NaOH) multiplied by 0.02688 L. Let's call this equation Y, just for laughs.
Now, let's substitute X into equation Y, so we have c(HClO4) * 0.025 L = X * 0.02688 L. I'm not sure about you, but I see a lot of L's in there. Perhaps we should go on a field trip to an alphabet factory.
But, let's not give up just yet, my friend! We can rearrange this equation to solve for c(HClO4). Dividing both sides of the equation by 0.025 L, we get c(HClO4) = X * 0.02688 L / 0.025 L.
Now, all we need to do is substitute X with c(OH) * 0.00413 L, which we defined earlier. So, c(HClO4) = c(OH) * 0.00413 L * 0.02688 L / 0.025 L. Phew, that was a wild ride!
To find the molar concentration, you just need to divide the moles by the volume of the solution. And since I'm a clown, let me tell you, I can divide things really well. Just kidding!
I hope that helps you with your calculations, and remember, when in doubt, add a dash of humor to the mix. Good luck with your chemistry adventures, my friend!
To calculate the molar concentrations of the acid (HClO4) and the base (NaOH), we need to follow the steps you've already started.
1. Calculate the molar concentration of Na2CO3:
- You already calculated the moles of Na2CO3 to be 0.0023068 mol.
- Given that you dissolved it in 50 mL of dilute perchloric acid solution, you can find the molar concentration:
c(Na2CO3) = n(Na2CO3) / V(acid)
where V(acid) is the volume of the acid in liters.
In this case, V(acid) = 50 mL = 0.050 L.
c(Na2CO3) = 0.0023068 mol / 0.050 L ≈ 0.046 mol/L.
2. Calculate the moles of HClO4 in the back titration:
- You mentioned that you used 4.13 mL of dilute NaOH in the back titration.
- Using this volume, you can calculate the moles of HClO4:
n(HClO4) = c(NaOH) * V(NaOH)
where c(NaOH) is the known concentration of the NaOH solution, and V(NaOH) is the volume in liters.
In this case, V(NaOH) = 0.00413 L (since you used 4.13 mL = 0.00413 L).
Since the moles of NaOH equal the moles of HClO4, we can also say:
n(HClO4) = n(NaOH).
3. Calculate the concentration of HClO4 using the separate titration:
- You mentioned that in a separate titration, a 25 mL portion of the acid required 26.88 mL of NaOH.
- From this information, we can write:
c(HClO4) * V(HClO4) = c(NaOH) * V(NaOH)
where V(HClO4) is the volume of HClO4 in liters.
In this case, V(HClO4) = 0.025 L (since you used 25 mL = 0.025 L).
Rearranging the equation:
c(HClO4) = (c(NaOH) * V(NaOH)) / V(HClO4)
Using the above equations and the given values, we can calculate the molar concentrations of the acid (HClO4) and the base (NaOH).