20.00 cm 3 of a solution containing 0.53 g of anhydrous Na2CO3 in 100 cm3 requires 25.00 cm3 of H2SO4 for complete neutralisation. The concentration of the acid solution in moles per dm is [H =1, C = 12, O = 16, Na =23, S = 32]
Na2CO3 + H2SO4 ==> CO2 + H2O + Na2SO4
mols Na2CO3 = 0.53/106 = 0.005
M Na2CO3 = mols/dm^3 = 0.005/0.1 = 0.05 M
mols Na2CO3 used in the titration = M x L = 0.05 M x 0.020 = 0.001
M H2SO4 = mols/dm^3 = 0.001/0.025 = 0.04 M
Well, let's do some chemistry calculations, shall we?
First, we need to find the number of moles of Na2CO3 in the given amount of the solution. To do that, we divide the mass of Na2CO3 (0.53 g) by its molar mass (2 * 23 + 12 + 3 * 16), which equals 105 g/mol. So, 0.53 g / 105 g/mol = 0.005 moles.
Now, we need to find the concentration of Na2CO3 in moles per dm3. Since we had 0.005 moles in 100 cm3, we can convert that to dm3 by dividing by 1000: 0.005 moles / 0.1 dm3 = 0.05 moles/dm3.
Finally, we want to find the concentration of H2SO4 in moles per dm3. Since 25.00 cm3 of H2SO4 were needed to neutralize the Na2CO3, we know that those 25.00 cm3 contain the same number of moles as the Na2CO3, which is 0.005 moles. So, the concentration is 0.005 moles / 0.025 dm3 = 0.2 moles/dm3.
Thus, the concentration of the acid solution in moles per dm3 is 0.2 moles/dm3.
To find the concentration of the H2SO4 solution in moles per dm3, we need to determine the number of moles of H2SO4 present in the 25.00 cm3 sample.
Given:
Volume of H2SO4 solution = 25.00 cm3
Concentration of Na2CO3 solution = 0.53 g/100 cm3
First, let's calculate the number of moles of Na2CO3:
Molar mass of Na2CO3 = (2*23 + 12 + 3*16) g/mol = 106 g/mol
Number of moles of Na2CO3 = mass / molar mass = 0.53 g / 106 g/mol = 0.005 moles
Since Na2CO3 and H2SO4 react in a 1:1 ratio, the number of moles of H2SO4 required for complete neutralization is also 0.005 moles.
Now, let's determine the concentration of H2SO4 in moles per dm3 (M):
Volume of H2SO4 solution = 25.00 cm3 = 0.025 dm3
Number of moles of H2SO4 = 0.005 moles
Concentration of H2SO4 = moles/volume = 0.005 moles / 0.025 dm3 = 0.2 moles/dm3
Therefore, the concentration of the H2SO4 solution is 0.2 moles/dm3.
To find the concentration of the acid solution in moles per dm³, we need to determine the number of moles of H₂SO₄ used in the neutralization reaction.
Step 1: Calculate the number of moles of Na₂CO₃ in the given solution.
The molar mass of Na₂CO₃ = (2 x 23) + 12 + (3 x 16) = 106 g/mol
Mass of Na₂CO₃ in the solution = 0.53 g
Number of moles of Na₂CO₃ = mass / molar mass
= 0.53 g / 106 g/mol
= 0.005 moles
Step 2: Use the balanced equation to find the mole ratio of Na₂CO₃ to H₂SO₄.
The balanced equation for the reaction is:
Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂
From the equation, the mole ratio of Na₂CO₃ to H₂SO₄ is 1:1.
Step 3: Calculate the number of moles of H₂SO₄ used.
Since the mole ratio is 1:1, the number of moles of H₂SO₄ is also 0.005 moles.
Step 4: Calculate the volume of the acid solution in dm³.
The volume of the acid solution used is 25.00 cm³.
To convert cm³ to dm³, divide by 1000:
Volume of acid solution = 25.00 cm³ / 1000
= 0.025 dm³
Step 5: Calculate the concentration of the acid solution in moles per dm³.
Concentration (in moles/dm³) = moles of H₂SO₄ / volume of solution in dm³
= 0.005 moles / 0.025 dm³
= 0.20 moles/dm³
Therefore, the concentration of the acid solution is 0.20 moles/dm³.