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A solution of 0.5moldm-3 NaOH was titrated against 0.2moldm-3 H2SO4 in the burette using phenolphthalein as indicator. The results obtained are: Initial Burette reading = 2.98 +0.05cm³ Final Burette reading = 38.75 10.05cm³ where ±0.05 is the maximum errors Determine the absolute error and relative error in ppt. of the measurements.​

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Final answer:

The total absolute error of the volume dispensed in the titration is 0.10 cm³, and the calculated relative error is approximately 2.8 parts per thousand (ppt).

Step-by-step explanation:

To calculate the absolute error and relative error of the measurements taken during a titration, you need to determine the error associated with the volume of the solution dispensed from the burette. In the given scenario, the initial burette reading is 2.98 ± 0.05 cm³ and the final burette reading is 38.75 ± 0.05 cm³. The absolute error for each reading is ± 0.05 cm³, so when considering the total volume dispensed, which is the difference between the final and initial readings, the errors must be added together.

The volume of H₂SO₄ dispensed is the final reading minus the initial reading: 38.75 cm³ - 2.98 cm³ = 35.77 cm³.

The total absolute error in the volume dispensed would be 0.05 cm³ + 0.05 cm³ = 0.10 cm³.

The relative error, often given in parts per thousand (ppt), is the absolute error divided by the total volume and then multiplied by 1,000:

Relative error (ppt) = (Absolute error / Total volume) x 1,000 = (0.10 cm³ / 35.77 cm³) x 1,000 ≈ 2.8 ppt.

This provides an indication of the precision of the measurements taken and is particularly important for achieving accurate results in analytical chemistry tasks such as titrations.

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