Friday

November 21, 2014

November 21, 2014

Posted by **Gloria** on Thursday, August 1, 2013 at 7:49am.

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 seperate 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.

- Analytical chemistry -
**Devron**, Thursday, August 1, 2013 at 12:08pmI 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

- Analytical chemistry -
**DrBob222**, Thursday, August 1, 2013 at 2:45pmI 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.

- Analytical chemistry -
**Devron**, Thursday, August 1, 2013 at 9:06pmI thought it was an easier way to do it, but it was super early/late and I couldn't reason it at the time.

- Analytical chemistry -
**Gloria**, Friday, August 2, 2013 at 7:25amThank you both so much

**Answer this Question**

**Related Questions**

Chemistry - How can i explain the key features of the periodic table relate to ...

Ap Chemistry - A perchloric acid solution is composed of 168.75 g of perchloric ...

College Chemistry - Sodium hydroxide is used extensively in acid-base titrations...

chemistry - You need to determine the concentration of a sulfuric acid solution ...

12th grade - A perchloric acid solution is composed of 168.75 g of perchloric ...

Chemistry Math - A 0.4671g sample containing sodium bicarbonate was titrated ...

analytical chemistry - An EDTA solution is standardized against a solution of ...

Chemistry - When titrating an acid with a base, a dilute solution of the base ...

Science - Are any of the bellow chemical changes reversible? If yes, then how? ...

chem - If 0.752 g of pure sodium carbonate was dissolved in water and the ...