[Rutherford]

I found recently two hard problems I don't have a clue how to solve them:

1.If in 200g of NaHSO4 solution( constantly mixing) is added 200g of Na2CO3 solution the mass of the obtained solution is 395,6g, but if the solutions get mixtured in reverse order the mass of the obtained solution is 397,8g. Calculate the w(%) of Na2CO3&NaHSO4 in the former solutions. (7,95% and 12%)-not a clue how to solve this one.

2.On balanced pans of a scale are two glasses with 100gHCl(5%). In first glass we add 10,82g of BaCO3. What mass of NaHCO3 we have to add in the second glass so that the pans are still balanced? (14,43g)-this one I actually tried a few times, but couldn't get the right result.

If someone could solve them and explain it to me, I would be really grateful.

[Borek]

If someone could solve them and explain it to me, I would be really grateful.

That would be against forum rules.

I guess the idea behind the first one is that when you add carbonate to hydrogen sulfate, there is a large excess of the H⁺ from the hydrogen sulfate, so carbonate gets protonated completely and carbonic acid immediately decomposes. If you add hydrogen sulfate to carbonate, carbonate is in a large excess, so it gets protonated to hydrogen carbonate which doesn't decompose. At least that's what may be happening initially. I am not convinced it would matter after mixing everything, as it will be a matter of equilibrium and how fast carbon dioxide leaves the solution, but perhaps I am wrong. If it works this way, it is a "Problem of the week" difficulty level question. Please tell us what was the solution after you will know the answer.

For the second - write both reaction equations. From the stoichiometry of the first reaction you should be able to calculate by how much mass of the beaker changed (you added some mass, some mass was lost to the environment). Use similar approach to calculate the reverse - how much sodium bicarbonate must be added for the same mass difference.

[Rutherford]

Oh, I am sorry for that. I will try to solve the 1st one like you said, but in the 2nd one I am not sure what mass is lost to the environment:
2HCl+BaCO3-->BaCl2+H2O+CO2
HCl+NaHCO3-->NaCl+H2O+CO2
Is it CO2 in both reactions?

[Borek]

Is it CO2 in both reactions?

Yes.

[Rutherford]

I finished a half of the 1st problem:
2NaHSO4+Na2CO3-->Na2SO4+H2O+CO2
The mass should be 200g+200g=400g, but it is 395,6g because of CO2. Mass of CO2 is 400g-395,6g=4,4g n(CO2)=4,4g/44(g/mol)=0,1mol, n(NaHSO4)=0,2mol, m(NaHSO4)=24g, w(NaHSO4)=24/200*100=12% and it is the same result like in the answer. Now I cant' finish the 2nd part:
Na2CO3+NaHSO4-->NaHCO3+Na2SO4 the mass should be again 400g but it is 397,8g, but why? No gass is formed.

And in the second one I got 17,64g but that is not the answer.

[Borek]

As I said earlier, first question is not clear to me, so I am not going to comment.

Second - beware, it is tricky. Do you have enough HCl to react with 17.64g of NaHCO₃?

[Rutherford]

I don't understand. I will write how I got the result and if you can, show me where I've mistaken:
In first glass is: 100g+10,82g-(10,82/208)*44=108,53g
In second glass: 100+84x-44x=108,53 x=0,213mol m=17,92g

[Borek]

How many grams of HCl would react with 17.92g NaHCO₃? How many grams of HCl are there in 100g of 5% w/w solution?

[Rutherford]

From 5g will react 7,79g (impossible). I tried this way:2nd glass:100+x(NaHCO3)-44*5/36,5=108,51 and I got 14,54g, 0,11g more than in the answer

[Borek]

0.11g can be just a difference in the molar masses used, or some rounding error.

[Rutherford]

Ok, thank you for helping me.