We can use stoichiometry to determine the volume of 0.01 M sodium carbonate solution required to neutralize 0.98 g of 1-¹ sulfuric acid in 25 cm³ of solution.
First, we need to calculate the number of moles of sulfuric acid present in the solution:
n(H₂SO4) = mass/molar mass = 0.98 g / 98 g/mol = 0.01 mol
From the balanced chemical equation, we can see that 1 mole of sodium carbonate (Na₂CO₃) reacts with 1 mole of sulfuric acid (H₂SO₄):
H₂SO4 (aq) + Na2CO3(aq) → Na₂SO4 (aq) + H₂O (l) + CO2 (g)
Therefore, the number of moles of sodium carbonate required to neutralize the sulfuric acid is also 0.01 mol.
We know that the concentration of the sodium carbonate solution is 0.01 M, which means that it contains 0.01 moles of Na₂CO₃ per liter of solution. To determine the volume of this solution required to neutralize the sulfuric acid, we can use the following formula:
V = n / C
where V is the volume of the sodium carbonate solution required (in liters), n is the number of moles of Na₂CO₃ required (which we calculated as 0.01 mol), and C is the concentration of the sodium carbonate solution (which is given as 0.01 M).
Plugging in the values, we get:
V = 0.01 mol / 0.01 M = 1.0 liter
However, this answer is in liters and we need to convert it to cm³, which is the same as milliliters (mL):
V = 1.0 liter x 1000 cm³/liter = 1000 cm³
Therefore, 1000 cm³ (or 1 liter) of 0.01 M sodium carbonate solution is required to neutralize 25 cm³ of a solution containing 0.98 g 1-¹ of sulfuric acid.