Final answer:
To reach the equivalence point in the titration of 40.0 mL of 0.100 M HNO₃, 28.57 mL of 0.140 M NaOH is required, calculated using the molar ratio and molarity of the solutions.
Step-by-step explanation:
The question involves calculating the volume of 0.140 M NaOH needed to reach the equivalence point in the titration of 40.0 mL of 0.100 M HNO₃. The neutralization reaction between NaOH and HNO₃ is a 1:1 molar ratio, since one mole of NaOH reacts with one mole of HNO₃ to form water and a salt. To find the volume of NaOH required, we can use the formula:
n₁(M₁V₁) = n₂(M₂V₂),
Where n₁ and n₂ are the stoichiometric coefficients (which are both 1), M₁ and M₂ are the molarities of HNO₃ and NaOH respectively, and V₁ and V₂ are the volumes of HNO₃ and NaOH respectively.
By rearranging and solving for V₂ (the volume of NaOH), we have:
V₂ = (M₁V₁) / M₂
V₂ = (0.100 M * 40.0 mL) / 0.140 M
V₂ = 28.57 mL
Therefore, 28.57 mL of 0.140 M NaOH is needed to reach the equivalence point in the titration of 40.0 mL of 0.100 M HNO₃.