[btx3]

1) A 2.0 L container holds 2.0 mole of HBr (which can decompose into H2 and Br2 in an equilibrium reaction). What will be the final concentrations of all species once equilibrium is reached? [Hint: the concentration of HBr is not 0.0 M] K = 6.2 * 10^-13.
Attempt: concentration OF HBr is 1 mol/L, do I just use k=([H2].[Br2])/([HBr])^2 to calculate the concentration of all species? I feel like there is a trick so can someone guide me
2) A 0.3240g sample of impure Na2CO3 was dissolved in 50.00mL of 0.1280M HCl. 50.00 mL of pure H2O was added. The excess acid then requires an average titre of 30.10mL of NaOH for complete neutralization. The NaOH was first standardised with oxalic acid (a diprotic acid). 25.00 mL of NaOH required 5.37, 5.42, 5.78 and 5.34 mL of acid to be neutralised (the titration was repeated 4 times. Calculate the % Na2CO3 (MM = 105.99) in the sample. Try to minimise any random or systematic errors in the data through your calculations.
Attempt:
Na2CO3 + HCl --> 2NaOH + H2O + CO2
When water us added to this solution, we get CO2 + H2O --> H2CO3, correct? How do we calculate the concentration of this acid?
Please help me!

[Borek]

1) A 2.0 L container holds 2.0 mole of HBr (which can decompose into H2 and Br2 in an equilibrium reaction). What will be the final concentrations of all species once equilibrium is reached? [Hint: the concentration of HBr is not 0.0 M] K = 6.2 * 10^-13.
Attempt: concentration OF HBr is 1 mol/L, do I just use k=([H2].[Br2])/([HBr])^2 to calculate the concentration of all species? I feel like there is a trick so can someone guide me

The correct, full blown approach would be to use ICE table. Your approach is not incorrect (although what you suggest is just a first step, and an incomplete one), but it requires understanding why it can be used.

2) A 0.3240g sample of impure Na2CO3 was dissolved in 50.00mL of 0.1280M HCl. 50.00 mL of pure H2O was added. The excess acid then requires an average titre of 30.10mL of NaOH for complete neutralization. The NaOH was first standardised with oxalic acid (a diprotic acid). 25.00 mL of NaOH required 5.37, 5.42, 5.78 and 5.34 mL of acid to be neutralised (the titration was repeated 4 times. Calculate the % Na2CO3 (MM = 105.99) in the sample. Try to minimise any random or systematic errors in the data through your calculations.
Attempt:
Na2CO3 + HCl --> 2NaOH + H2O + CO2

Huh? You added strong acid to produce a strong base? And what happened to Cl on the right?

When water us added to this solution, we get CO2 + H2O --> H2CO3, correct? How do we calculate the concentration of this acid?

No, carbon dioxide left the solution and it will not interfere with the further proceedings.