A 0.250 M NaOH solution was used to titrate a 19.75 mL HI solution. The endpoint was reached after 32.95 mL of titrant were delivered. Find the molar concentration of HI in the original solution.
To find the molar concentration of HI in the original solution, we can use the concept of stoichiometry. The balanced equation for the reaction between NaOH and HI is:
NaOH + HI -> NaI + H2O
From the balanced equation, we can see that the ratio between the moles of NaOH and HI is 1:1.
1. Calculate the number of moles of NaOH used in the titration.
To do this, we need to use the equation:
moles of NaOH = molarity x volume in liters
Given that the molarity of the NaOH solution is 0.250 M and the volume delivered is 32.95 mL (0.03295 L), we can calculate the moles of NaOH used:
moles of NaOH = 0.250 M x 0.03295 L = 0.0082375 moles
2. Calculate the number of moles of HI in the original solution.
Since the ratio between NaOH and HI is 1:1, the moles of NaOH used also represent the moles of HI in the original solution. Therefore, the moles of HI in the original solution is:
moles of HI = 0.0082375 moles
3. Calculate the molar concentration of HI in the original solution.
To find the molar concentration, we need to divide the moles of HI by the volume of the original solution. The volume of the original solution is given as 19.75 mL (0.01975 L). Therefore:
molar concentration of HI = moles of HI / volume of original solution
= 0.0082375 moles / 0.01975 L
= 0.417 M
Therefore, the molar concentration of HI in the original solution is 0.417 M.
To find the molar concentration of HI in the original solution, we can use the concept of stoichiometry and the volume of titrant used.
Here's how to solve the problem step by step:
1. Determine the balanced chemical equation for the reaction between NaOH and HI:
NaOH + HI -> NaI + H2O
2. Calculate the number of moles of NaOH used in the titration:
Moles of NaOH = Molarity of NaOH × Volume of NaOH used (in L)
Moles of NaOH = 0.250 M × (32.95 mL / 1000 mL/L) (Note: We convert mL to L by dividing by 1000 mL/L)
Moles of NaOH = 0.0082375 mol
3. Use the stoichiometry of the balanced equation to determine the number of moles of HI:
From the balanced equation, we can see that the mole ratio of NaOH to HI is 1:1. This means that the number of moles of HI is also 0.0082375 mol.
4. Calculate the molar concentration of HI in the original solution:
Molar concentration of HI = Moles of HI / Volume of HI used (in L)
Molar concentration of HI = 0.0082375 mol / (19.75 mL / 1000 mL/L) (Note: We convert mL to L by dividing by 1000 mL/L)
Molar concentration of HI = 0.417297 mol/L
Therefore, the molar concentration of HI in the original solution is 0.417297 mol/L.