Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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

Answer 5

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.