An HNO3 solution has a pH of 3.04. What volume of 0.015 M LiOH will be required to titrate 89.0 mL of the HNO3 solution to reach the equivalence point?
To find the volume of 0.015 M LiOH required to titrate the HNO3 solution, we need to use the concept of stoichiometry.
Step 1: Write the balanced chemical equation for the reaction between HNO3 and LiOH:
HNO3 + LiOH -> LiNO3 + H2O
From the balanced equation, we can see that one mole of HNO3 reacts with one mole of LiOH to produce one mole of LiNO3 and one mole of water.
Step 2: Calculate the number of moles of HNO3 in the given solution:
Moles of HNO3 = Molarity of HNO3 x Volume of HNO3 solution
Moles of HNO3 = 10^(-pH) x Volume of HNO3 solution
Given that pH = 3.04 and Volume of HNO3 solution = 89.0 mL, we can calculate the moles of HNO3:
Moles of HNO3 = 10^(-3.04) x 0.089 L
Step 3: Determine the volume of LiOH required to react with the moles of HNO3:
From the balanced chemical equation, we see that the mole ratio between HNO3 and LiOH is 1:1. This means that one mole of LiOH is needed to react with one mole of HNO3.
Volume of LiOH solution = Moles of HNO3 x (1 mole LiOH / 1 mole HNO3) / Molarity of LiOH
Given that the Molarity of LiOH is 0.015 M, we can substitute the values into the equation:
Volume of LiOH solution = (10^(-3.04) x 0.089 L) x (1 mole LiOH / 1 mole HNO3) / 0.015 M
Now, you can substitute the given values into the equation and perform the calculations to find the volume of LiOH solution required to titrate the HNO3 solution to the equivalence point.
HNO3 + LiOH ==> LiNO3 + H2O
The rxn is 1:1; therefore, we may use
mLHNO3 x MHNO3 = mLLiOH x MLiOH
Convert pH of 3.04 to H^+ by pH = -log(H^+) first, substitute, then solve for mL LiOH.