Question

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.

Answer 1

Answer:

The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.

Explanation:

The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

$C = \frac{n}{V} \\0.0275 = \frac{n}{0.0005}$

n = 1.375x10⁻⁵ 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):

1.375x10⁻⁵ mol _________ 19.9 units

        x              _________  25.2 units

x = 1.741x10⁻⁵mol

Finally, we can calculate the Cu²⁺ concentration :

C = 1.741x10⁻⁵mol / 0.025 L

C = 6.964x10⁻⁴ M