Answer :
35.00 mL of 0.737 M H2SO4 requires 62.4 mL of 0.827 M KOH solution for neutralization. So, the answer is D) 62.4.
To solve this problem, we can use the concept of stoichiometry from chemistry. In a neutralization reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH), the balanced chemical equation is:
H2SO4 + 2KOH → K2SO4 + 2H2O
This equation tells us that one mole of sulfuric acid reacts with two moles of potassium hydroxide. From the given information, we can calculate the number of moles of H2SO4:
moles of H2SO4 = (volume of H2SO4 solution in liters) × (molarity of H2SO4)
= (35.00 mL ÷ 1000 mL/L) × (0.737 mol/L)
≈ 0.0258 moles
According to the balanced equation, the mole ratio between H2SO4 and KOH is 1:2. Therefore, the number of moles of KOH required for neutralization is twice the number of moles of H2SO4:
moles of KOH = 2 × moles of H2SO4
≈ 2 × 0.0258
≈ 0.0516 moles
Now, we can use the definition of molarity to find the volume of KOH solution needed:
(volume of KOH solution in liters) = (moles of KOH) ÷ (molarity of KOH)
= 0.0516 moles ÷ 0.827 mol/L
≈ 0.0624 L
Converting this to milliliters:
0.0624 L × 1000 mL/L ≈ 62.4 mL
So, the correct answer is option D) 62.4.
The correct answer is D). In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.
To determine how many milliliters of 0.827 M KOH solution is required to neutralize 35.00 mL of 0.737 M H₂SO₄, we first need to consider the balanced chemical equation for the neutralization reaction:
H₂SO₄ (aq) + 2KOH (aq) → K₂SO₄ (aq) + 2H₂O (l)
From the equation, we see that 1 mole of H₂SO₄ reacts with 2 moles of KOH. Therefore, the mole ratio between H₂SO₄ and KOH is 1:2.
Let's calculate the moles of H₂SO₄ first:
Moles of H₂SO₄ = volume (in liters) × concentration
Moles of H₂SO₄ = 0.035 L × 0.737 mol/L
Moles of H₂SO₄ = 0.025795 mol
According to the mole ratio, 2 moles of KOH are required for every 1 mole of H₂SO₄. Therefore, moles of KOH required:
Moles of KOH = 2 × moles of H₂SO₄
Moles of KOH = 2 × 0.025795 mol
Moles of KOH = 0.05159 mol
Now, calculate the volume of 0.827 M KOH solution required:
Volume (in liters) of KOH = moles / concentration
Volume (in liters) of KOH = 0.05159 mol / 0.827 mol/L
Volume (in liters) of KOH = 0.0624 L
Convert liters to milliliters:
Volume of KOH = 0.0624 L × 1000 mL/L
Volume of KOH = 62.4 mL
Therefore, the correct answer is D) 62.4 mL of 0.827 M KOH solution is required for neutralization.
