4) Three unknown solutions were prepared by pipetting 15.00 m into a 200.00 mL v

Question

4) Three unknown solutions were prepared by pipetting 15.00 m into a 200.00 mL volumetric L of an unknown nickel solution flask and diluting to the mark. Then three samples were prepared from this dilute flask, to which the reagents were added then it was diluted to the mark. UV Spe analyzed the resulting solutions and the unknown absorbance value were con concentration by means of a calibration curve, Table 4-1. ill in the diagram volumes. Determine the concentration of the original nickel solution, in ppm Ni three unknown solutions and an average. Included appropriate error calculatio complete calculations using Trial 1 data. Present results for all trials and the average with nown solution by pipetting 5.00 mL of the dilute solution into a 25.00 ml verted to with appropriate ns. Show uncertainties (19 marks) Table 4-1 Unknown Trial Ni] (ppm) 2.768m ± 0.15252 3.23552± 0.16244 3.35263 ± 0.16485 3 foc V, C, V a52

Explanation / Answer

From the given data

Average Ni concentration from three readings

= (2.768 + 3.235 + 3.352)/3 = 3.118

Uncertainty

= (0.152 +.162 + 0.164)/3 = 0.159

So concentration of Ni in diluted 25 ml solution = 3.118 +/- 0.159 ppm

concentration in 5 ml aliquot solution,

= (3.118 x 25/5) +/- (0.159 x 25/5)

= 15.59 +/-0.795 ppm

Concentration of Ni in original 15 ml solution

= (15.59 x 200/15) +/- (0.795 x 200/15)

= 207.867 +/-10.600 ppm

Related Questions

Navigate

• Browse (All)
• Browse 4
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.