A is a solution of tetraoxosulphate (vi) acid, B is a solution containing 1.4g of potassium hydroxide per 250cm3.
using the table above as the value for your titration and 25cm3 for the value of pipette used
1. concentration of B in mol/dm3
2. concentration of A in mol/dm3 3. number of hydrogen ion in 1.0dm3 of A?
To find the answers to these questions, we first need to determine the number of moles of substance in each solution. After that, we can calculate the concentration in moles per liter (mol/dm³) using the given volumes.
1. Concentration of B in mol/dm³:
Given that the mass of potassium hydroxide (KOH) in solution B is 1.4g and the volume is 250cm³, we can find the number of moles (n) using the formula:
n = mass / molar mass
The molar mass of KOH is:
K = 39.10 g/mol (potassium)
O = 16.00 g/mol (oxygen)
H = 1.01 g/mol (hydrogen)
Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol
n = 1.4g / 56.11 g/mol ≈ 0.025 mol
Now, let's calculate the concentration:
Concentration (mol/dm³) = moles / volume (dm³)
Given that the volume of solution B is 250cm³, which is equal to 0.250 dm³:
Concentration (mol/dm³) = 0.025 mol / 0.250 dm³ = 0.10 mol/dm³
Therefore, the concentration of solution B is 0.10 mol/dm³.
2. Concentration of A in mol/dm³:
To find the concentration of solution A, we need to perform a titration. During the titration, a known volume of solution A is reacted with a known volume of solution B until the reaction is complete. The idea is to determine the volume of B required to neutralize A completely.
Since the table is mentioned but not provided, we will assume the stoichiometry of the reaction and calculate the concentration based on the balanced chemical equation:
H₂SO₄ + 2KOH ⟶ K₂SO₄ + 2H₂O
The stoichiometry of the reaction implies that one mole of H₂SO₄ reacts with two moles of KOH.
Given that the volume used for pipetting solution A is 25 cm³, which is equal to 0.025 dm³, and if the reaction is complete, it will react with an equal volume of solution B.
Therefore, the moles of KOH in solution B that react with solution A are equal to the moles of A.
Concentration (mol/dm³) = moles / volume (dm³) = 0.025 mol / 0.025 dm³ = 1.00 mol/dm³
Hence, the concentration of solution A is 1.00 mol/dm³.
3. Number of hydrogen ions in 1.0 dm³ of A:
From the balanced chemical equation:
2H₂SO₄ ⟶ K₂SO₄ + 2H₂O
We observe that for every one mole of H₂SO₄, there are two moles of hydrogen ions (H⁺).
Therefore, for 1.0 dm³ (1000 cm³) of solution A with a concentration of 1.00 mol/dm³, the number of hydrogen ions is:
Number of hydrogen ions = concentration × volume (dm³)
Number of hydrogen ions = 1.00 mol/dm³ × 1.0 dm³ = 1.00 mol
Hence, there are 1.00 mol of hydrogen ions in 1.0 dm³ of solution A.
To determine the concentration of B in mol/dm^3, we need to calculate the number of moles of potassium hydroxide (KOH) in 1.0 dm^3 of solution.
Given that there is 1.4 g of KOH in 250 cm^3 of solution, we can use this to find the molar mass of KOH:
Molar mass of KOH = Mass / Moles
Molar mass of KOH = 1.4 g / Moles
The molar mass of KOH is approximately 56.1 g/mol.
Now, let's convert the volume from cm^3 to dm^3:
Volume = 250 cm^3 = 250/1000 dm^3 = 0.25 dm^3
To calculate the concentration of B in mol/dm^3, we'll use the formula:
Concentration (mol/dm^3) = Moles / Volume (dm^3)
Since the mass of KOH is given, we can calculate the number of moles using the formula:
Moles = Mass / Molar mass
Plugging in the values, we have:
Moles = 1.4 g / 56.1 g/mol = 0.0249 mol
Now, we can calculate the concentration:
Concentration of B = 0.0249 mol / 0.25 dm^3 = 0.0996 mol/dm^3
Therefore, the concentration of solution B is 0.0996 mol/dm^3.
To determine the concentration of A in mol/dm^3, we need to perform a titration.
Unfortunately, the "table above" you mentioned is not provided, so we cannot use the values from the table to determine the concentration of A.
Please provide the necessary information from the table in order to proceed with the calculations.
Regarding the number of hydrogen ions in 1.0 dm^3 of solution A, we need to determine the number of moles of hydrogen ions present.
Since the formula for tetraoxosulphate (VI) acid is H2SO4, we know that for every molecule of H2SO4, there are 2 hydrogen ions (H+).
Given the concentration of solution A, we can calculate the number of moles of hydrogen ions using the formula:
Moles = Concentration (mol/dm^3) * Volume (dm^3)
If you provide the concentration of A and the volume used (in dm^3), we can calculate the number of moles of hydrogen ions and consequently the number of hydrogen ions in 1.0 dm^3 of solution A.
I don't see a table.
1. (KOH) = mols/L
mols = 1.4 g/molar mass = ?
Then mols/dm3 = M = ?mols from above/0.250 L = ?
Did you use 25 cc H2SO4? How much KOH was used in the titration?