In a titration, how many millilitres of 0.23 mol/L NaOH(aq) must be added to 11 mL of 0.18 mol/L HI(aq) to reach the equivalence point?
Write the equation.
Calculate mols HCl present
Determine mols NaOH needed.
Calculate mL NaOH required.
Post your work if you get stuck.
To solve this problem, you can use the concept of stoichiometry and the balanced equation for the reaction between NaOH and HI:
NaOH(aq) + HI(aq) -> NaI(aq) + H2O(l)
Since the balanced equation has a 1:1 molar ratio between NaOH and HI, the number of moles of NaOH needed will be equal to the number of moles of HI present.
Step 1: Calculate the number of moles of HI:
moles of HI = concentration of HI × volume of HI
moles of HI = 0.18 mol/L × 11 mL
moles of HI = 0.018 mol
Step 2: Determine the volume of NaOH needed:
moles of NaOH = moles of HI
concentration of NaOH × volume of NaOH = moles of HI
0.23 mol/L × volume of NaOH = 0.018 mol
volume of NaOH = 0.018 mol / 0.23 mol/L
volume of NaOH ≈ 0.078 mL
Therefore, approximately 0.078 mL of 0.23 mol/L NaOH(aq) must be added to 11 mL of 0.18 mol/L HI(aq) to reach the equivalence point.
To answer this question, you need to use the concept of stoichiometry and the equation of the reaction between HI and NaOH. The balanced equation for the reaction is:
HI(aq) + NaOH(aq) → H2O(l) + NaI(aq)
From the balanced equation, you can see that the stoichiometric ratio between HI and NaOH is 1:1. This means that 1 mole of HI reacts with 1 mole of NaOH.
To determine the amount of NaOH required to reach the equivalence point, you can use the equation:
n (moles) = C (concentration, mol/L) x V (volume, L)
First, calculate the number of moles of HI in the initial solution:
moles of HI = C HI x V HI
moles of HI = 0.18 mol/L x 0.011 L
moles of HI = 0.00198 mol
Since the stoichiometric ratio between HI and NaOH is 1:1, you know that the number of moles of NaOH needed is also 0.00198 mol.
Now, use the equation to calculate the volume of NaOH required:
V NaOH = n NaOH / C NaOH
V NaOH = 0.00198 mol / 0.23 mol/L
V NaOH ≈ 0.0861 L
Convert the volume to millilitres:
V NaOH ≈ 0.0861 L x 1000 mL/L
V NaOH ≈ 86.1 mL
Therefore, approximately 86.1 millilitres of 0.23 mol/L NaOH(aq) must be added to 11 mL of 0.18 mol/L HI(aq) to reach the equivalence point.