Question

There is 25.3 mL of the sulfuric acid solution; the sulfuric acid concentration is 2.30 M. Your base solution is 1.00 M. What is the volume in mL of base that is required to complete the neutralization of the acid

Answer 1

Answer:

The volume of the base is 0.05819·x L, where x is the number of moles of base that combines with one mole of sulfuric acid

Explanation:

The volume of the sulfuric acid, V = 25.3 mL = 25.3 × 10⁻³ L

The concentration of the sulfuric acid, c = 2.30 M

The concentration of the base, $c_{base}$ = 1.00 M

Let the mole ratio of the acid to base be 1 : x

The number of moles of sulfuric acid present, n = c × V

∴ n = 2.30 M/L × 25.3 × 10⁻³ L = 0.05819 moles

The number of moles of sulfuric acid present, n = 0.05819 moles

1 mole of sulfuric acid combines with x moles of base

Therefore, 0.05819 moles of sulfuric acid will combine with 0.05819·x moles of base

The number of moles of base, $n_{base}$ = 0.05819·x moles

Therefore, the volume of base, $V_{base}$ = $n_{base}$/$c_{base}$

∴  $V_{base}$ = 0.05819·x/1 ≈ 0.05819·x L

The volume of base,  $V_{base}$ ≈ 0.05819·x L.