Question

1. 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.

Answer 1

6.83x10⁻⁴M is the initial concentration of the Cu²⁺

Step-by-step explanation:

The concentration is directly proportional to the signal (Beer-Lambert law).

The increase in concentration produce an increase in signal of:

45.1 units - 25.2 units = 19.9 units.

The increase in concentration is of:

0.500mL = 5x10⁻⁴L * (0.0275mol / L) = 1.375x10⁻⁵moles / 0.0255L = 5.39x10⁻⁴M

The conversion factor is 5.39x10⁻⁴M = 19.9 units

That means the initial concentration that produce a signal of 25.2 units is:

25.2 units * (5.39x10⁻⁴M / 19.9 units) =

6.83x10⁻⁴M is the initial concentration of the Cu²⁺