[Rutherford]

Some sample of NaOH absorbed CO₂ and partially transformed into Na₂CO₃. 1g of that sample was dissolved in water so that the volume of the solution gets 100cm³. 20cm³ of that solution were titrated with HCl (c=0.2mol/dm³) with phenolphthalein and 24.22cm³ of the HCl solution were spent. Another 20cm³ of the beginning solution were titrated with the same HCl solution with methyl orange and the solution was heated so that CO₂ dissapears from it. 24.7cm³ of HCl were spent. Calculate the mass share of Na₂CO₃ in the sample.

Tried many times to solve, but I couldn't understand this:
1.Is there NaHCO₃ in the sample?
2.Why were different quantities of HCl spent? Is it related to the heating?
3.Why were different indicators used?

[Borek]

http://www.titrations.info/acid-base-titration-sodium-hydroxide-and-carbonate

[Rutherford]

Ok, there are NaOH, NaHCO₃ and Na₂CO₃ in the sample. If I understood well: OH^- will react with HCl first and it can't be detected with an indicator. Then CO₃^2- reacts which can be seen because of phenolphthalein, so 24.22cm³ of the HCl solution were spent on these two. When using methyl orange, after the previous two, HCO₃^- react with HCl, and now are 24.7cm³ of HCl spent. V₁-V₂ is spent on HCO3^-, but there is said that it is spent on CO₃^2-. Don't understand that part.

[Borek]

Ok, there are NaOH, NaHCO₃ and Na₂CO₃ in the sample.

Can NaOH coexist with NaHCO₃?

[Rutherford]

Is V₂-V₁ then used for Na₂CO₃?

[Borek]

Yes. That's what is written on the the page I linked to, isn't it?

[Rutherford]

Yes but the part with NaHCO₃ confused me. Got the right result (5%). So, in similar problems I have to know what indicator is used for what base in the sample. Thanks again.

[Rutherford]

Managed to do it, but after few hours, when I tried to do it again I got 2 times smaller result  . Don't know where I've mistaken, so I need help.
Calculated the n of HCl that was used to titrate Na₂CO₃
n(HCl)=(0.0247-0.02422)*0.2=9.6*10^-5mol. n(Na₂CO₃)=n(HCl)/2=4.8*10^-5mol.
c of Na2CO3 is the same in the 20cm³ and in the 100cm³. c=4.8*10^-5/0.02=2.4*10^-3mol/dm³.
n of Na₂CO₃ in 100cm³ is n=2.4*10^-3*0.1=2.4*10^-4mol
m=2.4*10^-4*106=0.025g or 2.5%. The right answer is 5%.

[Borek]

n(Na₂CO₃)=n(HCl)/2=4.8*10^-5mol

Why do you divide by two? What is the reaction equation?

[Rutherford]

I think that I know what you mean. I used this equation:
Na₂CO₃+2HCl 2NaCl+H₂O+CO₂, but the correct one would be:
Na₂CO₃+HCl NaHCO₃+NaCl so it's not divided by 2. Right?

[Rutherford]

After a lot of thinking and by the help of the link you gave me I think that I understood now. I am sorry for being so dumb, but I think this problem is a little too hard for just one year of studying chemistry in high school.