contestada

A 25.0-mL sample containing Cu2+ gave an instrument signal of 25.2 units (corrected for a blank). When exactly 0.500 mL of 0.0275 M Cu(NO3)2 was added to the solution, the signal increased to 45.1 units. Calculate the molar concentration of Cu2+ assuming that the signal was directly proportional to the analyte concentration. Skoog, Douglas A.. Principles of Instrumental Analysis (p. 20). Brooks Cole. Kindle Edition.

Respuesta :

fgschem

Answer:

The molar concentration of CuĀ²āŗ in the initial solution is 6.964x10ā»ā“ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NOā‚ƒ)ā‚‚ added to the solution:

[tex]C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005} \\[/tex]

n = 1.375x10ā»āµ mol

The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):

1.375x10ā»āµ mol _________ 19.9 units

Ā  Ā  Ā  Ā  x Ā  Ā  Ā  Ā  Ā  Ā  Ā _________ Ā 25.2 units

x = 1.741x10ā»āµmol

Finally, we can calculate the CuĀ²āŗ concentration :

C = 1.741x10ā»āµmol / 0.025 L

C = 6.964x10ā»ā“ M

Otras preguntas