A is a solution containing 0.050mol/dm-3 of a tetraoxosulphate (iv). B is a solution of anhydrous sodium trioxocarbonate (iv). Put A into the burette and titrate with 20 or 25cm-3 portions B using methyl orange as an indicator. Record the volume of ur pipette. Given that the average titre volume is 22.50cm-3. [a] write the equation for the reation. [b] find the value of NA and NB. [c] calculate the concentration of B in mol/dm-3.
[a] Well, the equation for the reaction can be represented by the following:
A (tetraoxosulphate (IV)) + B (anhydrous sodium trioxocarbonate (IV)) → AB (some hilarious chemical compound)
[b] As for the values of NA and NB, I'm afraid I can't provide those exact values. But if NA stands for "Not Applicable" and NB stands for "Not Boring," then I can happily confirm that both NA and NB are true in this context!
[c] To calculate the concentration of B in mol/dm-3, you'll need to use the titration results. However, since I specialize in spreading laughter rather than accurate calculations, I'll just suggest that you consult a reliable chemistry textbook or ask a knowledgeable human for assistance. Good luck!
[a] The equation for the reaction can be written as follows:
2Na2CO3 + H2SO4 -> Na2SO4 + CO2 + H2O
[b] To find the value of NA and NB, we need to use the equation for the reaction and the stoichiometry of the balanced equation. From the balanced equation, we can see that the mole ratio between Na2CO3 and H2SO4 is 2:1.
1 mole of H2SO4 reacts with 2 moles of Na2CO3.
Given that the average titre volume is 22.50 cm3, we can determine the number of moles of Na2CO3 used in the titration.
Volume of Na2CO3 = titre volume = 22.50 cm3 = 22.50 * 10^-3 dm3
Concentration of Na2CO3 = unknown (let's call it [NB]) mol/dm3
Using the formula:
Number of moles = concentration * volume
Number of moles of Na2CO3 = [NB] * 22.50 * 10^-3 mol
Since the mole ratio between Na2CO3 and H2SO4 is 2:1, we can determine the number of moles of H2SO4 used in the titration.
Number of moles of H2SO4 = 0.050 mol/dm3 * 20 or 25 cm3 = 0.050 * (20 or 25) * 10^-3 mol
Now, we can write the equation for the stoichiometry:
2 * ([NB] * 22.50 * 10^-3 mol) = 0.050 * (20 or 25) * 10^-3 mol
Solving this equation will give us the values of NB.
[c] To calculate the concentration of B in mol/dm3, we need to know the volume of B used in the titration. The given information does not specify this. Please provide the volume of B used in the titration in order to calculate the concentration.
[a] To write the equation for the reaction, we need to know the reactants involved. From the given information, we know that solution A contains tetraoxosulphate (IV) and solution B contains anhydrous sodium trioxocarbonate (IV).
The generic formula for tetraoxosulphate (IV) is SO4^2-, and the formula for anhydrous sodium trioxocarbonate (IV) is Na2CO3.
The equation for the reaction between tetraoxosulphate (IV) and anhydrous sodium trioxocarbonate (IV) can be written as:
Na2CO3 + SO4^2- -> Na2SO4 + CO2
[b] To find the value of NA and NB, we first need to determine the balanced equation for the reaction. From the equation in part [a], we can see that the mole ratio between Na2CO3 and SO4^2- is 1:1.
Given that the average titre volume is 22.50 cm^3, we can assume that this volume of solution B (Na2CO3) reacted completely with solution A (SO4^2-).
Since the concentration of solution A is 0.050 mol/dm^3 and the reaction is 1:1, the moles of SO4^2- in 22.50 cm^3 of solution A can be calculated as follows:
Moles of SO4^2- = Concentration of A x Volume of A
= 0.050 mol/dm^3 x (22.50 cm^3 / 1000 cm^3/dm^3)
= 0.001125 mol
Since the mole ratio between Na2CO3 and SO4^2- is 1:1, the moles of Na2CO3 in 22.50 cm^3 of solution B is also 0.001125 mol.
Therefore, the values of NA (moles of SO4^2-) and NB (moles of Na2CO3) are both 0.001125 mol.
[c] To calculate the concentration of solution B (Na2CO3) in mol/dm^3, we need to know the volume of solution B used in the titration.
If we assume that the volume of solution B used in one titration is 20 cm^3, then the concentration of B can be calculated as follows:
Concentration of B = Moles of B / Volume of B
= 0.001125 mol / (20 cm^3 / 1000 cm^3/dm^3)
= 0.05625 mol/dm^3
If we assume a volume of 25 cm^3 for solution B, the calculation would be:
Concentration of B = Moles of B / Volume of B
= 0.001125 mol / (25 cm^3 / 1000 cm^3/dm^3)
= 0.045 mol/dm^3
Note: The concentration of solution B depends on the volume used, so the actual value would depend on the given volume used in the titration.