[TimTim]

Ok. A titration is carried out to determine the percentage of sodium hydroxide and sodium carbonate in a mixture solution. Basically, this titration is between 0.1M dilute hydrochloric acid with a MIXTURE of unkown concentration which contains sodium hydroxide and sodium carbonate.

In Part 1, the MIXTURE is titrated directly with dilute hydrochloric acid.

In Part 2 , the MIXTURE is added with barium chloride, BaCl_2.

I was told that in Part 2, BaCl_2 can remove sodium carbonate from the mixture solution by forming insoluble barium carbonate.

My question is:
1. How does BaCl_2 remove sodium carbonate?
2. Will BaCl_2 react with NaOH and affects percentage volume of NaOH?
3. What is the reactions occur in Part 1 ?
4. What is the reactions occur in Part 2 ?

Edited : Typos

[Borek]

Please read forum rules.

No such thing as BaCl.

You were told correctly.

What do you know about barium and calcium carbonates solubility?

[TimTim]

It is actually the typos I made and corrected right after u made your post. ( Besides that it is a homework, I am frustrated for not solving this even after spending 2 days google around )

I will start the beginning by myself.

Before titration started:

Na₂CO₃ + 2 HCl -> 2NaCl + H₂O + CO₂

NaOH + HCl -> NaCl + H₂O

If Barium Chloride is added....

I know that : Na₂CO₃ + BaCl₂ -> NaCl + BaCO₃

But will ? : BaCl₂ + NaOH -> Ba(OH)₂ + 2NaCl 
                Ba(OH)₂ + CO₂ -> BaCO₃ + H₂O
                Volume of NaOH affected??!?!

What do you know about barium and calcium carbonates solubility?

Barium carbonate is insoluble.

[Borek]

Are you sure you were not told to filtrate precipitate?

What do you mean by NaOH volume - if anything, you are interested in HCl volume.

[TimTim]

Are you sure you were not told to filtrate precipitate?

What do you mean by NaOH volume - if anything, you are interested in HCl volume.

I am not told to filtrate the precipitate. According to my manual, I am supposed to titrate the mixture solution and measure the volume of HCl required for titration.

According to the lecturer, the concentration of NaOH is obtained by using the volume of HCl required in Part 2. She says Concentration of NaOH can be obtained directly since Na₂CO₃is removed.

I use the formula n₁/n₂= M₁V₁/M₂V₂

About the 2nd portion of your question, I don't get it. My english level sucks. I know that BaCl₂ can remove Na₂CO₃. Just that I wonder will it affect the amount of NaOH in the solution ? If it is affected, the concentration of NaOH that I obtain is wrong.

I am new to this course and limited classmates and resources... Please help out >< !

[Borek]

OK, so you are asking about AMOUNT of NaOH, not the volume. BaCl₂ doesn't matter - it doesn't react woth NaOH. Or even if it does - precipitating Ba(OH)₂ - precipitate will easily dissolve in HCl and amount if OH^- determined will be the same.

However, what makes me wonder is the idea that it is enough to precipitate BaCO₃ to get rid of carbonate. IMHO it will easily dissolve in the presence of HCl, so if it is not filtrated, it still adds to the volume of HCl used.

[TimTim]

I see. that is why I get an inaccurate result. It uses extra HCl to neautralise the mixture. Maybe there is stil carbonate left there.

By the way, can I ask why BaCl₂ will not react with NaOH?

[Borek]

By the way, can I ask why BaCl₂ will not react with NaOH?

There are no thermodynamically preferred products of such reaction.

[TimTim]

By the way, can I ask why BaCl₂ will not react with NaOH?

There are no thermodynamically preferred products of such reaction.

You mean.. Na₂CO₃ is more preferred to be reacted with BaCl₂ compared with NaOH ?
hehe... What actually meant by 'no thermodynamically preferred products' ? >< sorry i'm weak.

[Borek]

Things react when products are for some reason more stable then reactants. In the case of BaCl₂ and NaOH possible products (NaCl and Ba(OH)₂) are just as stable as the original substances. In the case of BaCl₂ and Na₂CO₃ BaCO₃ precipitates, as it is more stable in the solid form than in the solution.

Note that 'stability' I am using here is a layman's term, without precise definition.

[TimTim]

I see. Thanks for your explanation and guiding me when I'm "lost".

I guess I can partially understand now. Imma look through reference books in the library and google about stability thing. Thanks a lot. Borek rocks