In preparing a standard alkaline solution, a learner dissolve 14,0 g of potassium hydroxide in 1,00 dm^3 of water. In a titration, 30,0cm^3 of the KOH(aq) is neutralised by a sulfuric acid solution with a concentration of 0,100 mol•dm^-3. Calculate the concentration of the alkaline solution. And calculate the volume of H2SO4( aq) used in the titration.
To find the concentration of the alkaline solution, we can first determine the number of moles of potassium hydroxide (KOH) in the solution.
1. Convert the mass of KOH to moles:
mass of KOH = 14.0 g
molar mass of KOH = 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol
moles of KOH = mass / molar mass = 14.0 g / 56.11 g/mol ≈ 0.249 mol
2. Calculate the volume of the alkaline solution:
volume of alkaline solution = 1.00 dm³ = 1000 cm³
3. Calculate the concentration of the alkaline solution:
concentration of KOH solution = moles of KOH / volume of alkaline solution
concentration = 0.249 mol / 1000 cm³ ≈ 0.249 mol/dm³
So, the concentration of the alkaline solution is approximately 0.249 mol/dm³.
To find the volume of sulfuric acid (H2SO4) used in the titration, we can use the stoichiometry of the balanced chemical equation.
The balanced equation for the neutralization reaction between KOH and H2SO4 is:
2 KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2 H2O(l)
From the balanced equation, we can see that 2 moles of KOH react with 1 mole of H2SO4.
4. Calculate the number of moles of H2SO4 used in the titration:
moles of H2SO4 = concentration of H2SO4 × volume of H2SO4
moles of H2SO4 = 0.100 mol/dm³ × 30.0 cm³ × (1 dm³/1000 cm³) = 0.003 mol
5. Use the stoichiometry of the balanced equation to find the volume of H2SO4 used:
moles of H2SO4 = 0.003 mol
moles of KOH = 2 × 0.003 mol = 0.006 mol
volume of H2SO4 = moles of KOH / concentration of KOH
volume of H2SO4 = 0.006 mol / 0.249 mol/dm³ ≈ 0.024 dm³
So, the volume of H2SO4 used in the titration is approximately 0.024 dm³ or 24 cm³.
Bot, that is incorrect.
M KOH is correct @ 0.249 mols/dm^3.
H2SO4 + 2KOH --> K2SO4 + 2H2O
mols KOH used = M x L = 0.249 x 0.030 L = 0.00749
mols H2SO4 = 0.00749 moles KOH x (1 mol H2SO4/2 moles NaOH) = 0.003735
M H2SO4 = moles H2SO4/dm^3 H2SO4 or
dm^3 H2SO4 = moles H2SO4/M H2SO4 = 0.003735/0.1 = 0.03735 dm^3 = 37.35 cm^3 = volume H2SO4 used.
Apologies for the incorrect response earlier. You're absolutely right.
To calculate the concentration of the alkaline solution:
The concentration of KOH is indeed 0.249 mol/dm³.
To calculate the volume of H2SO4 used in the titration:
Given:
Moles of KOH used = 0.00749 mol (calculated using M x L)
Moles of H2SO4 = 0.00749 mol KOH x (1 mol H2SO4/2 mol KOH) = 0.003745 mol
Concentration of H2SO4 = 0.100 mol/dm³
Volume of H2SO4 used = Moles of H2SO4 / Concentration of H2SO4
Volume of H2SO4 used = 0.003745 mol / 0.100 mol/dm³ = 0.03745 dm³ = 37.45 cm³
So, the concentration of the alkaline solution is 0.249 mol/dm³, and the volume of H2SO4 used in the titration is 37.45 cm³.
From your statement:
Moles of KOH used = 0.00749 mol (calculated using M x L).
so M = 0.249 and L = 0.030 = 0.00747 and not 0.00749
Apologies for the error in my previous response. You are absolutely correct.
Using the correct calculation:
Moles of KOH used = M x L = 0.249 mol/dm³ x 0.030 dm³ = 0.00747 mol
Moles of H2SO4 = 0.00747 mol KOH x (1 mol H2SO4/2 mol KOH) = 0.00374 mol
Concentration of H2SO4 = 0.100 mol/dm³
Volume of H2SO4 used = Moles of H2SO4 / Concentration of H2SO4
Volume of H2SO4 used = 0.00374 mol / 0.100 mol/dm³ = 0.0374 dm³ = 37.4 cm³
So, the concentration of the alkaline solution is 0.249 mol/dm³, and the volume of H2SO4 used in the titration is 37.4 cm³.
To calculate the concentration of the alkaline solution, we can use the balanced equation for the neutralization reaction between potassium hydroxide (KOH) and sulfuric acid (H2SO4):
2 KOH + H2SO4 -> K2SO4 + 2 H2O
From the equation, we can see that 2 moles of KOH react with 1 mole of H2SO4. Given that the concentration of the sulfuric acid solution is 0.100 mol/dm^3 (or 0.100 M), and 30.0 cm^3 (or 0.030 dm^3) is used in the titration, we can calculate the number of moles of H2SO4 used:
Moles of H2SO4 = Concentration x Volume
= 0.100 mol/dm^3 x 0.030 dm^3
= 0.003 mol
Since the stoichiometric ratio of KOH to H2SO4 is 2:1, the number of moles of KOH neutralized is also 0.003 mol.
Now, let's calculate the molar mass of KOH:
- The atomic mass of potassium (K) is 39.10 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
- The atomic mass of hydrogen (H) is 1.01 g/mol.
Hence, the molar mass of KOH = 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol.
Using the given mass of KOH (14.0 g) and the molar mass, we can calculate the number of moles of KOH:
Moles of KOH = Mass / Molar mass
= 14.0 g / 56.11 g/mol
= 0.249 mol
The concentration of the alkaline solution is given by dividing the moles of KOH by the volume of the solution in dm^3:
Concentration of KOH solution = Moles of KOH / Volume of solution
= 0.249 mol / 1.00 dm^3
= 0.249 mol/dm^3 (or 0.249 M)
Therefore, the concentration of the alkaline solution is 0.249 mol/dm^3 (or 0.249 M).
Now to calculate the volume of H2SO4 used in the titration, we can use the stoichiometry of the reaction:
2 moles of KOH reacts with 1 mole of H2SO4
From earlier calculations, we know that 0.003 mol of H2SO4 is used. Therefore, the number of moles of KOH used would be half of that:
Moles of KOH used = 0.003 mol / 2
= 0.0015 mol
Using the concentration of the KOH solution (0.249 mol/dm^3) and the equation Moles = Concentration x Volume, we can calculate the volume of KOH solution used:
Volume of KOH solution used = Moles of KOH used / Concentration of KOH solution
= 0.0015 mol / 0.249 mol/dm^3
= 0.006 dm^3 (or 6.0 cm^3)
Therefore, the volume of H2SO4 solution used in the titration is 6.0 cm^3 (or 0.006 dm^3).
To calculate the concentration of the alkaline solution, we need to use the concept of moles and volume.
Step 1: Convert the mass of potassium hydroxide (KOH) to moles.
Given that the mass of KOH is 14.0 g, we first need to determine the molar mass of KOH.
The molar mass of KOH is:
(1 * molar mass of K) + (1 * molar mass of O) + (1 * molar mass of H)
= (1 * 39.1 g/mol) + (1 * 16.0 g/mol) + (1 * 1.01 g/mol)
= 56.11 g/mol
Now, we can calculate the number of moles of KOH by dividing the mass of KOH by its molar mass:
Number of moles of KOH = 14.0 g / 56.11 g/mol
= 0.2490 mol
Step 2: Calculate the concentration of the alkaline solution.
Given that the volume of the solution is 1.00 dm^3, we can use the formula:
Concentration = Number of moles / Volume
Concentration of the alkaline solution = 0.2490 mol / 1.00 dm^3
= 0.2490 mol/dm^3
Therefore, the concentration of the alkaline solution is 0.2490 mol/dm^3.
Step 3: Calculate the volume of H2SO4(aq) used in titration.
Given that 30.0 cm^3 of KOH(aq) is neutralized by the sulfuric acid solution, we can use stoichiometry to determine the volume of H2SO4(aq) used.
The balanced chemical equation is:
2 KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2 H2O(l)
The stoichiometry of the balanced equation indicates that 2 moles of KOH react with 1 mole of H2SO4.
Since we have the concentration of H2SO4 as 0.100 mol/dm^3, we can calculate the number of moles of H2SO4 used as follows:
Number of moles of H2SO4 = Concentration * Volume
Number of moles of H2SO4 = 0.100 mol/dm^3 * 0.0300 dm^3
Number of moles of H2SO4 = 0.00300 mol
Since the stoichiometric ratio is 2:1, the number of moles of KOH used is also 0.00300 mol.
Now we need to convert the number of moles of KOH used to volume. Using the same formula as before:
Volume of H2SO4(aq) used = Number of moles / Concentration
Volume of H2SO4(aq) used = 0.00300 mol / 0.100 mol/dm^3
Volume of H2SO4(aq) used = 0.0300 dm^3
Therefore, the volume of H2SO4(aq) used in the titration is 0.0300 dm^3 (or 30.0 cm^3).