Solution B was formed by dissolving 1.06g of Na2CO3 in 250 cm^3 of solution. Solution A was prepared from a concentrated HCl ( density = 1.18 kg/dm^3, Assay = 36-37% (m/m)) by measuring 0.25 cm^3 of the concentrated HCl in 100 cm^3 of solution. Find the volume of the diluted HCl that is needed to react completely with the Na2CO3
What's the molarity of the HCl solution? Note 1.18 kg/dm^3 = 1.18 g/mL. Also the molarity of the HCl can't be calculated exactly so I'll take the average of 36-37% as the assay. That means the EXACT volume of HCl needed to react with the Na2CO3 isn't known but the value calculated is close.
That's 1000 mL x 1.18 g/mL x 0.365 x (1/36.5) = 11.6 M for the solution A.
Molarity of the diluted HCl solution is 11.6 x (0.25 cc/100 cc) = 0.029 M
Molarity of the Na2CO3 solution.
mols Na2CO3 = grams/molar mass = 1.06/106 = 0.01 mols
M Na2CO3 = mols/L = 0.01/0.250 = 0.04 M
2HCl + Na2CO3 = 2NaCl + H2O + CO2
millimols Na2CO3 = mL x M = 250 mL x 0.04 M = 10
It will take 20 mmols HCl. Look at the coefficients to know that.
M diluted HCl = mmols/mL
0.029M = 20 mmols/mL
Solve for mL = ? but that number looks unrealistic to me. Check out the numbers.
To determine the volume of diluted HCl required to react completely with Na2CO3, we need to calculate the amount of Na2CO3 and HCl involved in the reaction, and then use stoichiometry to find the volume of HCl. Here's how we can do that step by step:
1. Calculate the number of moles of Na2CO3:
- Given mass of Na2CO3: 1.06g
- Molar mass of Na2CO3: 22.99 g/mol (Na) + 12.01 g/mol (C) + (3 * 16.00 g/mol) (O) = 105.99 g/mol
- Number of moles of Na2CO3 = mass / molar mass = 1.06g / 105.99 g/mol = 0.00999 mol (approximately 0.01 mol)
2. Determine the balanced chemical equation between Na2CO3 and HCl:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
1 mol of Na2CO3 reacts with 2 mol of HCl.
3. Calculate the number of moles of HCl needed to react with Na2CO3:
From the balanced equation, we know that 1 mol of Na2CO3 reacts with 2 mol of HCl. Therefore, 0.01 mol of Na2CO3 will react with 0.01 mol * (2 mol HCl / 1 mol Na2CO3) = 0.02 mol of HCl.
4. Calculate the volume of concentrated HCl required to obtain 0.02 mol:
- Given volume of concentrated HCl: 0.25 cm^3
- Convert cm^3 to dm^3 by dividing by 1000: 0.25 cm^3 / 1000 = 0.00025 dm^3
- Concentration of HCl in dm^3: 1.18 kg/dm^3 (density of HCl)
- Mass of HCl in the given volume: mass = density * volume = 1.18 kg/dm^3 * 0.00025 dm^3 = 0.000295 kg
- Assay is the percentage (m/m) of the desired substance in the solution, in this case, HCl. The assay is 36-37% (m/m), so we take the average: (36% + 37%) / 2 = 36.5%. This means that 100 g of the solution contains 36.5 g of HCl.
- Calculate the amount of HCl in the given volume: amount = (percentage / 100) * mass = (36.5 / 100) * 0.000295 kg = 0.000107675 kg = 0.107675 g
- Convert grams to moles of HCl: moles = mass / molar mass = 0.107675 g / 36.46 g/mol (molar mass of HCl) = 0.0029538 mol (approximately 0.003 mol)
5. Determine the volume of diluted HCl needed to react completely with Na2CO3:
From the stoichiometry of the reaction, we know that 1 mol of HCl reacts with 1 mol of Na2CO3. Therefore, 0.003 mol of HCl will react with 0.003 mol of Na2CO3.
But we have already calculated that 0.01 mol of Na2CO3 is present.
Therefore, we need 0.01 mol / 0.003 mol HCl = 3.33... times more HCl than we currently have.
So, the volume of diluted HCl needed is 3.33... times the initial volume of HCl:
3.33... * 0.25 cm^3 = 0.833... cm^3
Therefore, approximately 0.834 cm^3 of the diluted HCl is needed to react completely with the Na2CO3.