Mass total – mass in sample \frac{0.003108 \text{ mol} \times 55.85 \text{g}}{1 \text{mol}} = 0.1736 \text{ g} \times 100 (0.36 \text{g} – 0.1736 \text{g}) The student learns that the actual mass percent of iron in the ore sample is greater than the experimental value calculated in part (C). The student claims that the difference between the actual and experimental values could have been caused by rinsing the buret with distilled water just prior to filling it with 0.017 M K
Cr
O
(aq). Do you agree with the student’s claim? Explain your reasoning. Disagree, the residual distilled water would dilute the iron and cause the actual mass to be less. higher
The Correct Answer and Explanation is:
Answer:
Disagree. The student’s claim is incorrect. Rinsing the buret with distilled water just prior to filling it with 0.017 M K₂Cr₂O₇ (aq) would dilute the titrant, not the iron solution. This dilution would cause a lower concentration of the titrant, leading to an overestimation of the volume of titrant needed to reach the endpoint, and thus an underestimation of the amount of iron in the sample. This explains why the experimental mass percent of iron is lower than the actual value.
Explanation
In a redox titration involving potassium dichromate (K₂Cr₂O₇) and iron (Fe²⁺), accurate determination of iron content depends on precise titrant concentration and volume measurements. If the buret is rinsed with distilled water before being filled with the potassium dichromate solution, the residual water left in the buret will dilute the titrant as it is added. This results in a lower effective concentration of the titrant than intended.
Because the titrant is now less concentrated, a larger volume must be dispensed to fully react with the same amount of iron ions. However, the student typically calculates the number of moles of iron based on the assumed (undiluted) concentration of the dichromate solution. This leads to the underestimation of iron moles in the sample.
Consequently, when calculating the mass of iron using: mass Fe=mol Fe×55.85 g/mol\text{mass Fe} = \text{mol Fe} \times 55.85\,\text{g/mol}mass Fe=mol Fe×55.85g/mol
the underestimated moles result in a smaller calculated iron mass. When this mass is divided by the mass of the original ore sample and multiplied by 100 to find the percent by mass, the result is lower than the actual value.
Therefore, while the discrepancy between the actual and experimental values is valid, the reasoning provided by the student is flawed. The dilution affects the titrant, not the analyte, and leads to underestimation, not overestimation, of the iron content. The correct explanation aligns with the observed result: the experimental value being lower than the true value.
