In an experiment 0.0500 M AgNO3 was used to titrate 20 mL of 0.12 M NaCl solution. When
silver electrode is used together with the pH electrode to measure potential at each point during
the titration, the potential measured at equivalence point is 0.055 V. Unfortunately, a student
doing the titration stopped at –0.055 V. Calculate the relative titration error (in percentage) of the
student’s result.
To calculate the relative titration error, we need to compare the student's result to the correct result at the equivalence point.
Given:
Initial concentration of AgNO3 = 0.0500 M
Volume of AgNO3 used = 20 mL = 0.020 L
Concentration of NaCl = 0.12 M
First, let's determine the number of moles of AgNO3 used in the titration:
moles of AgNO3 = concentration x volume
moles of AgNO3 = 0.0500 M x 0.020 L
moles of AgNO3 = 0.001 mol
According to the balanced chemical equation:
AgNO3 + NaCl → AgCl + NaNO3
The stoichiometry of the reaction tells us that 1 mol of AgNO3 reacts with 1 mol of NaCl to form 1 mol of AgCl.
Therefore, the number of moles of NaCl in the original solution is also 0.001 mol.
Now, let's find the volume of NaCl in the original solution:
moles of NaCl = concentration x volume
0.001 mol = 0.12 M x volume of NaCl
volume of NaCl = 0.001 mol / 0.12 M
volume of NaCl = 0.00833 L = 8.33 mL
At the equivalence point, the reaction between AgNO3 and NaCl is complete, and all the Ag+ ions have reacted to form AgCl. The potential measured at the equivalence point should be 0 V.
However, the student stopped the titration at -0.055 V.
Relative Titration Error = (|Measured Potential - Expected Potential| / Expected Potential) x 100%
Relative Titration Error = (|-0.055 V - 0 V| / 0 V) x 100%
Relative Titration Error = (0.055 V / 0 V) x 100%
Relative Titration Error = infinity
The relative titration error is infinity, indicating that the student's result is significantly different from the expected result at the equivalence point.
To calculate the relative titration error, we need to compare the student's result with the expected result at the equivalence point.
Step 1: Calculate the number of moles of AgNO3 used.
Molarity (M) = moles/volume in liters
0.0500 M AgNO3 * (20 mL / 1000 mL) = 0.001 moles AgNO3
Step 2: Determine the stoichiometric ratio between AgNO3 and NaCl.
From the balanced chemical equation, we know that 1 mole of AgNO3 reacts with 1 mole of NaCl.
Step 3: Calculate the number of moles of NaCl present in the 20 mL solution.
moles NaCl = moles AgNO3 = 0.001 moles
Step 4: Calculate the concentration of NaCl in the 20 mL solution.
Molarity (M) = moles/volume in liters
0.001 moles / (20 mL / 1000 mL) = 0.050 M NaCl
Step 5: Calculate the theoretical volume of AgNO3 needed at the equivalence point.
Using the stoichiometry, we know that 1 mole of AgNO3 reacts with 1 mole of NaCl. Therefore, at the equivalence point, the number of moles of AgNO3 used should be equal to the number of moles of NaCl present in the 20 mL solution.
Step 6: Calculate the volume of AgNO3 needed at the equivalence point.
Molarity (M) = moles/volume in liters
0.12 M NaCl * volume = 0.001 moles NaCl
volume = 0.001 moles NaCl / 0.12 M NaCl = 0.00833 L = 8.33 mL
Step 7: Calculate the relative titration error in percentage.
Relative titration error = |student's result - expected result| / expected result * 100%
= |(-0.055 V) - (0.055 V)| / (0.055 V) * 100%
= 0.11 V / 0.055 V * 100%
= 200%
Therefore, the relative titration error of the student's result is 200%.