A25.0-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.

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:

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