Final answer:
To neutralize 90.0 mL of 0.205 M LiOH, 92.25 mL of 0.200 M HNO₃ is required based on the 1:1 stoichiometry of their neutralization reaction.
Step-by-step explanation:
To calculate the volume of 0.200 M HNO₃ needed to neutralize 90.0 mL of 0.205 M LiOH, a stoichiometric calculation can be used based on the reaction:
HNO₃(aq) + LiOH(aq) → LiNO₃(aq) + H₂O(l)
For a 1:1 molar ratio between HNO₃ and LiOH, the moles of HNO₃ needed will be equal to the moles of LiOH. First, calculate the moles of LiOH:
Moles of LiOH = Volume of LiOH × Molarity of LiOH = 0.0900 L × 0.205 mol/L = 0.01845 moles
Using the mole ratio from the balanced chemical equation (1 mole of HNO₃ per mole of LiOH), the same number of moles of HNO₃ are required to neutralize the LiOH.
Now find the volume of 0.200 M HNO₃:
Volume of HNO₃ = Moles of HNO₃ / Molarity of HNO₃ = 0.01845 moles / 0.200 mol/L = 0.09225 L
Convert liters to milliliters:
Volume of HNO₃ in mL = 0.09225 L × 1000 mL/L = 92.25 mL
Therefore, 92.25 mL of 0.200 M HNO₃ is required to neutralize 90.0 mL of 0.205 M LiOH